|log{xx_{1}}| + |log{xx_{2}}| + …+ |log{xx_{n}}| + |log{x/x_{1}}| + |log{x/x_{2}}| + …+|log{x/x_{n}}|= |log{x_{1}}+ log{x_{2}}+ ….+log{x_{n}}|

Solve the following :

Find all positive real numbers $x, x_{1}, x_{2}, \ldots, x_{n}$ such that

$|\log{xx_{1}}|+|\log{xx_{2}}| + \ldots + |\log{xx_{n}}| + |\log{\frac{x}{x_{1}}}| + |\log{\frac{x}{x_{2}}}| + \ldots + |\log{\frac{x}{x_{n}}}|= |\log{x_{1}}+ \log{x_{2}}+\log{x_{3}}+ \ldots + \log{x_{n}}|$
…let us say this is given equality A

Solution:

Use the following inequality: $|a-b| \leq |a| + |b|$ with equality iff $ab \leq 0$

So, we observe that : $|\log{xx_{1}}|+|\log{\frac{x}{x_{1}}}| \geq |\log{xx_{1}}-\log{\frac{x}{x_{1}}}| = |\log{x_{1}^{2}}|=2 |\log{x_{1}}|$,

Hence, LHS of the given equality is greater than or equal to:

$2(|\log{x_{1}}|+|\log{x_{2}}|+|\log{x_{3}}|+ \ldots + |\log{x_{n}}|)$

Now, let us consider the RHS of the given equality A:

we have to use the following standard result:

$|\pm a_{} \pm a_{2} \pm a_{3} \ldots \pm a_{n}| \leq |a_{1}|+|a_{2}| + \ldots + |a_{n}|$

So, applying the above to RHS of A:

$|\log{x_{1}}+\log{x_{2}}+\ldots + \log{x_{n}}| \leq |\log{x_{1}}|+|\log{x_{2}}|+\ldots + |\log{x_{n}}|$.

But, RHS is equal to LHS as given in A:

That is, $|\log{xx_{1}}|+|\log{xx_{2}}|+ \ldots + |\log{xx_{n}}| +|\log{\frac{x}{x_{1}}}|+|\log{\frac{x}{x_{2}}}|+ \ldots + |\log{\frac{x}{x_{n}}}| \leq |\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|$

Now, just a few steps before we proved that LHS is also greater than or equal to : That is,

$|\log{xx_{1}}|+|\log{xx_{2}}|+\ldots + |\log{xx_{n}}|+ |\log{\frac{x}{x_{1}}}|+|\log{\frac{x}{x_{2}}}| + \ldots + |\log{\frac{x}{x_{n}}}| \geq 2(|\log{x_{1}}|+|\log{x_{2}}|+\ldots + |\log{x_{n}}|)$

The above two inequalities are like the following: $x \leq y$ and $x \geq 2y$; so what is the conclusion? The first inequality means $x2y$ or $x=2y$; clearly it means the only valid solution is $x=2y$.

Using the above brief result, we have here:

$|\log{x_{1}}|+|\log{x_{2}}|+ \ldots +|\log{x_{n}}| =2(|\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|)$

Hence, we get $|\log{x_{1}}|+|\log{x_{2}}|+ \ldots + |\log{x_{n}}|=0$, which in turn means that (by applying the definition of absolute value):

$|\log{x_{1}}|=|\log{x_{2}}|= \ldots =|\log{x_{n}}|$, which implies that $x_{1}=x_{2}= \ldots x_{n}=1$.

Substituting these values in the given logarithmic absolute value equation, we get:

$n \times |\log{x}|+ n \times |\log{x}|=0$, that is $2n \times |\log{x}|=0$, and as $n \neq 0$, this implies that $|\log{x}|=0$ which in turn means $x=1$ also.

Every function can be written as a sum of an even and an odd function

Let $f(x)$ be any well-defined function.

We want to express it as a sum of an even function and an odd function.

Let us define two other functions as follows:

$F(x) = \frac{f(x)+f(-x)}{2}$ and $G(x)=\frac{f(x)-f(-x)}{2}$.

Claim I: F(x) is an even function.

Proof I; Since by definition $F(x)= \frac{f(x)+f(-x)}{2}$, so $F(-x) = \frac{f(-x) +f(-(-x))}{2}=\frac{f(-x)+f(x)}{2} \Longrightarrow F(x) = F(-x)$ so that F(x) is indeed an even function.

Claim 2: G(x) is an odd function.

Proof 2: Since by definition $G(x) = \frac{f(x)-f(-x)}{2}$, so $G(-x) = \frac{f(-x)-f(-(-x))}{2} = \frac{f(-x)-f(x)}{2} = -\frac{f(x)-f(-x)}{2} = -G(-x) \Longrightarrow G(x) = -G(-x)$ so that G(x) is indeed an odd function.

Claim 3: $f(x)= F(x) + G(x)$

Proof 3: $F(x)+ G(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2} = \frac{f(x)}{2} + \frac{f(-x)}{2} + \frac{f(x)}{2} - \frac{f(-x)}{2} = f(x)$ indeed.

Why do we need proofs? In other words, difference between a mathematician, physicist and a layman

Yes, I think it is a very nice question, which kids ask me. Why do we need proofs? Well, here is a detailed explanation (I am not mentioning the reference I use here lest it may intimidate my young enthusiastic, hard working students or readers. In other words, the explanation is not my own; I do not claim credit for this…In other words, I am just sharing what I am reading in the book…)

Here it goes:

What exactly is the difference between a mathematician, a physicist, and a layman? Let us suppose that they all start measuring the angles of hundreds of triangles of various shapes, find the sum in each case and keep a record. Suppose the layman finds that with one or two exceptions, the sum in each case comes out to be 180 degrees. He will ignore the exceptions and say “the sum of the three angles in a triangle  is 180 degrees.” A physicist will be more cautious in dealing with the exceptional cases. He will examine them more carefully. If he finds that the sum in them is somewhere between 179 degrees to 180 degrees, say, then he will attribute the deviation to experimental errors. He will then state a law: The sum of three angles of any triangle is 180 degrees. He will then watch happily as the rest of the world puts his law to test and finds that it holds good in thousands of different cases, until somebody comes up with a triangle in which the law fails miserably. The physicist now has to withdraw his law altogether or else to replace it by some other law which holds good in all cases tried. Even this new law may have to be modified at a later date. And, this will continue without end.

A mathematician will be the fussiest of all. If there is even a single exception he will refrain from saying anything. Even when millions of triangles are tried without a single exception, he will not state it as a theorem that the sum of the three angles in ANY triangle is 180 degrees. The reason is that there are infinitely many different types of triangles. To generalize from a million to infinity is as baseless to a mathematician as to generalize from one to a million. He will at the most make a conjecture and say that there is a strong evidence suggesting that the conjecture is true. But that is not the same thing as a proving a theorem. The only proof acceptable to a mathematician is the one which follows from earlier theorems by sheer logical implications (that is, statements of the form : If P, then Q). For example, such a proof follows easily from the theorem that an external angle of a triangle is the sum of the other two internal angles.

The approach taken by the layman or the physicist is known as the inductive approach whereas the mathematician’s approach is called the deductive approach. In the former, we make a few observations and generalize. In the latter, we deduce from something which is already proven. Of course, a question can be raised as to on what basis this supporting theorem is proved. The answer will be some other theorem. But then the same question can be asked about the other theorem. Eventually, a stage is reached where a certain statement cannot be proved from any other earlier proved statement(s) and must, therefore, be taken for granted to be true. Such a statement is known as an axiom or a postulate. Each branch of math has its own axioms or postulates. For examples, one of the axioms of geometry is that through two distinct points, there passes exactly one line. The whole beautiful structure of geometry is based on 5 or 6 axioms such as this one. Every theorem in plane geometry or Euclid’s Geometry can be ultimately deduced from these axioms.

PS: One of the most famous American presidents, Abraham Lincoln had read, understood and solved all of Euclid’s books (The Elements) by burning mid-night oil, night after night, to “sharpen his mental faculties”. And, of course, there is another famous story (true story) of how Albert Einstein as a very young boy got completely “addicted” to math by reading Euclid’s proof of why three medians of a triangle are concurrent…(you can Google up, of course).

Regards,

Nalin Pithwa

Number theory : A set of friendly examples

Even and odd numbers

Two whole numbers are added together. If their sum is odd, which statements below are
always true? Which are always false? Which are sometimes true and sometimes false?
1 Their quotient is not a whole number.
2 Their product is even.
3 Their difference is even.
4 Their product is more than their sum.
5 If 1 is added to one of the numbers and the product is found, it will be even.
The Collatz conjecture

If it is odd, multiply it by 3 and then add 1.
If it is even, divide it by 2. Then repeat this process on the number just obtained. Keep repeating the procedure.

For example, if you start with 58, the resulting chain of numbers is
58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, …

The Collatz conjecture, made by Lothar Collatz in 1937, claims that, if you repeat this
process over and over, starting with any whole number greater than zero, eventually you will finish up with the sequence … 4, 2, 1, 4, 2, 1.

A conjecture is a statement that is thought to be true but has not been proved mathematically to be true for all cases. Although the Collatz conjecture has been shown to work − often very quickly − for many whole numbers, there are some quite small numbers that take a very long time to come down to … 4, 2, 1, 4, 2, 1.Apply this process to all the whole numbers greater than zero and less than or equal to 30.
For each one, find:
• how many steps it takes to reach 1 the frst time
• the largest number in the sequence. (For the sequence above, 58 takes 19 steps and reaches a maximum of 88.)
Look for shortcuts and work with a partner if you like.

Long division
Here is a way to check how good your long division skills are. If you are able to follow it
through and get to the end without making a mistake, you can consider yourself a qualiꏨed long division champion.
• Start with any two-digit number (for example, 58). Write it three times so that a six-digit  number is formed (585 858).

• Divide this number by 21. There should not be any remainder. If there is, try and find out where you made your mistake and fix it.
• Now divide this new four- or possibly five-digit number by 37. Once again, there should be no remainder.
• Finally, divide this number − which should by now have only three or four digits − by 13.
You will know if you got it right by looking at the number you are left with.
Explain why this exercise works.
(Doing any of this exercise on a calculator is still interesting but is de뀠nitely wimping out!)

Totient numbers

1 A totient number is the number of fractions between 0 and 1 (not including 0 or 1) for
a given denominator that cannot be reduced to a simpler equivalent fraction. The totient
number of 2 is 1, since we have $\frac{1}{2}$; of 3 it is 2, since we have $\frac{1}{3}$ and $\frac{2}{3}$; and of 4 it is also 2, since we have $\frac{1}{4}$
and $\frac{3}{4}$ ($\frac{2}{4}$ can be reduced to $\frac{1}{2}$). The totient number of 5 is 4, since we have $\frac{1}{5}$, $\frac{2}{5}$, $\frac{3}{5}$, $\frac{4}{5}$; and of 6 it is 2, since we have $\frac{1}{5}$ and
$\frac{5}{6}$. Find the totient numbers forall denominators up to 12.

2 For any denominator n, there are n fractions between 0 and 1 (including 0 but not 1). Of
these fractions, some will be counted towards the totient number of n, but others will
cancel down and count towards the totient number of one of the factors of n. Using this
information and the totient numbers from the previous question, calculate the totient
numbers for 15, 18, 20 and 24.

3 The totient number is related to the prime factors of the original number, since these will
determine which fractions can be cancelled. Using this information, calculate the totient numbers of 72, 81, 98 and 100.

$\bf{Last \hspace{0.1in}Digits \hspace{0.1in} of \hspace{0.1in}powers}$
$\bf{Square \hspace{0.1in}Numbers}$

Without using a calculator, can you say which of this set of numbers could not be square numbers?

8116801, 251301659, 3186842, 20720704.

You can just by checking the last digit (units digit) of each number.

Do a bit of experimentation with a calculator and find the four digits that square numbers end in. (This eliminates the third number in this set).

Now check out the pairs of digits that your odd square numbers end with. What digits are possible in the tens position of an odd square number? (This number eliminates the second number in the set).

Complete these sentences with what you have discovered:

* In a square number, the last digit (units digit) can only be _____, _______, _______, _______, _______ or _______.
* The second last digit (tens place) of an odd square number is always _______.

$\bf{Cube \hspace {0.1in} Numbers}$

Cube numbers behave rather differently.

A bit of experimentation will show that cube numbers can end in ANY digit (units place). This digit depends on the last digit (units place) of the original number being cubed.

Complete this table:
$\left| \begin{array}{cc} \mbox {if a number ends in} & \mbox{its cube will end in}\\ 0 & \\ 1 & \\ 2 & \\ 3 & \\ 4 & \\ 5 & \\ 6 & \\ 7 & \\ 8 & \\ 9 & \end{array}\right|$

$\bf{Fourth \hspace{0.1in}Powers}$

Fourth powers are in fact just square numbers that have been squared. For example, $7^{4}=7^{2} \times 7^{2}= 49^{2}=2401$.

Since $4^{2}=16$ and $9^{2}=81$, the last digit of a fourth power can only be 0, 1, 5 or 6.

$\bf{Fifth \hspace{0.1in}powers}$

Fifth powers have a magic of their own. Do a bit of experimentation to find out what it is. In p, particular, I suggest you create tables of second powers, third powers, fourth powers and fifth powers of all numbers from zero to 20. Check, compare…actually, it is fun to “compare rate of growth of powers with increasing integers”…this idea involves rudimentary ideas of calculus…

$\bf{Obstinate \hspace{0.1in} numbers}$

An odd number can usually be written as a sum of a prime number and another number, which is a power of two. This is true for all odd numbers greater than one but less than 100.

For example, if we choose 23, we can say that it is equal to $23=19 + 2^{2}$. There is one more way to represent 23: it is $7 + 2^{4}$. So, there are two ways to represent 23 as a sum of a prime number and a power of two. But, $21+2^{1}$ and $15+2^{3}$ do not work as both 21 and 15 are not prime numbers.

Some odd numbers like this can be expressed in many ways.

Try to find various representations as sum(s) of prime number and a power(s) of two of the following integers: 45, 29, 59, 95.

If you are adventurous or courageous, try to find such representations of all odd numbers lying from 1 to 100. You need a lot of patience and stamina and grit…but you will develop an “intuitive feel or tactile feel for numbers”…that’s the way math begins…

There are some odd numbers which cannot be expressed as a sum of a prime number and a power of two. Such numbers are called $\bf{obstinate \hspace{0.1in} numbers}$.

An example of an obstinate number is $\bf{251}$ as the working below shows:

$251-2^{1}=249=3 \times 83$

$251-2^{2}=247=13 \times 19$

$251-2^{3}=243=3 \times 81$

$251-2^{4}=235= 5 \times 47$

$251-2^{5}=219= 3 \times 73$

$251-2^{6}= 187= 11 \times 17$

$251-2^{7}=123=3 \times 41$

The next power of 2 is $2^{8}=256$, which is clearly greater than 251. Hence, 251 is an obstinate number.

In fact, 251 is the third obstinate number. The first two lie between 100 and 150. Find these two odd numbers keeping track of how you eliminated the other twenty three odd numbers between 100 and 150.

$\it{Remember \hspace{0.1in} to \hspace{0.1in} be \hspace{0.1in} systematic \hspace{0.1in}}$.

Making a list of the powers of two up to $2^{8}$ might be a good place to start with. Look for short cuts and patterns as you proceed further.

Have fun with numbers !!

Regards,
Nalin Pithwa.

Method of undetermined coefficients for PreRMO, PRMO and IITJEE Foundation maths

1. Find out when the expression $x^{3}+px^{2}+qx+r$ is exactly divisible by $x^{2}+ax+b$

Solution 1:

Let $x^{3}+px^{2}+qx+r=(x^{2}+ax+b)(Ax+B)$ where A and B are to be determined in terms of p, q, r, a and b. We can assume so because we know from the fundamental theorem of algebra that the if the LHS has to be of degree three in x, the remaining factor in RHS has to be linear in x.

So, expanding out the RHS of above, we get:

$x^{3}+px^{2}+qx+r=Ax^{3}+aAx^{2}+bAx+Bx^{2}+Bax+bB$

$x^{3}+px^{3}+qx+r=Ax^{3}+(aA+B)x^{2}+x(bA+aB)+bB$

We are saying that the above is true for all values of x: hence, coefficients of like powers of x on LHS and RHS are same; we equate them and get a system of equations:

$A=1$

$p=aA+B$

$bA+aB=q$

$bB=r$

Hence, we get $p=a+\frac{r}{b}$ and $bp-ba=r$ or that $b(p-a)=r$

Also, $b+aB=q$ so that $q=b+\frac{ar}{b}$ which means $q-b=\frac{a}{b}r$

but $\frac{r}{b}=B=p-a$ and hence, $q-b=\frac{a}{b}(p-a)$

So, the required conditions are $b(p-a)=r$ and $q-b=\frac{a}{b}(p-a)$.

2) Find the condition that $x^{2}+px+q$ may be a perfect square.

Solution 2:

Let $x^{2}+px+q=(Ax+B)^{2}$ where A and B are to be determined in terms of p and q; finally, we obtain the relationship required between p and q for the above requirement.

$x^{2}+px+q=A^{2}x^{2}+B^{2}+2ABx$ which is true for all real values of x;

Hence, $A^{2}=1$ so $A=1$ or $A=-1$

Also, $B^{2}=q$ and hence, $B=\sqrt{q}$ or $B=-\sqrt{q}$

Also, $2AB=p$ so that $2\sqrt{q}=p$ so $q=\frac{p^{2}}{4}$, which is the required condition.

3) To prove that $x^{4}+px^{3}+qx^{2}+rx+s$ is a perfect square if $(q-\frac{p^{2}}{4})^{2}=4s$ and $r^{2}=p^{2}s$.

Proof 3:

Let $x^{4}+px^{3}+qx^{2}+rx+s=(Ax^{2}+Bx+C)^{2}$

$x^{4}+px^{3}+qx^{2}+rx+s=A^{2}x^{4}+B^{2}x^{2}+C^{2}+2ABx^{3}+2BCx+2ACx^{2}$

$A^{2}=1$

$2AB=p$

$q=B^{2}+2AC$

$2BC=r$

$C^{2}=s$

$A=1$ or $A=-1$

$2AB=p \longrightarrow 2B=p \longrightarrow B=\frac{p}{2}$

$q=B^{2}+2AC=\frac{p^{2}}{4}+2\times \sqrt{s} \longrightarrow (q-\frac{p^{2}}{4})^{2}=4s$

$2 \times \frac{p}{2} \times \sqrt{s}=r \longrightarrow r^{2}=p^{2}s$

More later,

Nalin Pithwa.

PS: Note in the method of undetermined coefficients, we create an identity expression which is true for all real values of x.

Discuss the following “proof” of the (false) theorem:

If n is any positive integer and S is a set containing exactly n real numbers, then all the numbers in S are equal:

PROOF BY INDUCTION:

Step 1:

If $n=1$, the result is evident.

Step 2: By the induction hypothesis the result is true when $n=k$; we must prove that it is correct when $n=k+1$. Let S be any set containing exactly $k+1$ real numbers and denote these real numbers by $a_{1}, a_{2}, a_{3}, \ldots, a_{k}, a_{k+1}$. If we omit $a_{k+1}$ from this list, we obtain exactly k numbers $a_{1}, a_{2}, \ldots, a_{k}$; by induction hypothesis these numbers are all equal:

$a_{1}=a_{2}= \ldots = a_{k}$.

If we omit $a_{1}$ from the list of numbers in S, we again obtain exactly k numbers $a_{2}, \ldots, a_{k}, a_{k+1}$; by the induction hypothesis these numbers are all equal:

$a_{2}=a_{3}=\ldots = a_{k}=a_{k+1}$.

It follows easily that all $k+1$ numbers in S are equal.

*************************************************************************************

Regards,

Nalin Pithwa

Miscellaneous Algebra: pRMO, IITJEE foundation maths 2019

For the following tutorial problems, it helps to know/remember/understand/apply the following identities (in addition to all other standard/famous identities you learn in high school maths):

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

By the way, I hope you also know how to derive the above.Let me mention two methods to derive the above :

Method I: Using polynomial division in three variable, divide the dividend $a^{3}+b^{3}+c^{3}-3abc$ by the divisor $a+b+c$.

Method II: Assume that $P(X)$ is a polynomial with roots a, b and c. So, we know by the fundamental theorem of algebra that $P(X)=(X-a)(X-b)(X-c)$. Now, we also know that a, b and c satisfy P(X). Now, proceed further and complete the proof.

Let us now work on the tutorial problems below:

1) If $2s=a+b+c$, prove that $\frac{1}{s-a} + \frac{1}{s-b} + \frac{1}{s-c} = \frac{abc}{s(s-a)(s-b)(s-c)}$

2) If $x^{2}+a^{2}=2(xy+yz+zu-y^{2}-z^{2})$, prove that $x=y=z=u$.

Prove the following identities:

3) $b(x^{3}+a^{3})+ax(x^{2}-a^{2})+a^{3}(x+a)=(a+b)(x+a)(x^{2}-ax+a^{2})$

4) $(ax+by)^{2}+(ay-bx)^{2}+c^{2}x^{2}+c^{2}y^{2}=(x^{2}+y^{2})(a^{2}+b^{2}+c^{2})$

5) $(x+y)^{3}+ 3(x+y)^{2}z+3(x+y)z^{2}+z^{3}=(x+z)^{3}+3(x+z)^{2}y+3(x+z)y^{2}+y^{3}$

6) $(a+b+c)(ab+bc+ca)-abc=(a+b)(b+c)(c+a)$

7) $(a+b+c)^{2}-a(b+c-a)-b(a+c-b)-c(a+b-c)=2(a^{2}+b^{2}+c^{2})$

8) $(x-y)^{3}+(x+y)^{3}+3(x-y)^{2}(x+y)+3(x+y)^{2}(x-y)=8x^{3}$

9) $x^{2}(y-z)+y^{2}(z-x)+z^{2}(x-y)+(y-z)(z-x)(z-y)=0$

10) $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)=-(b-c)(c-a)(a-b)(a+b+c)$

11) Prove that $(b-c)^{3}+(c-a)^{3}+(a-b)^{3}=3(b-c)(c-a)(a-b)$

12) If3 $2s=a+b+c$, prove that $(s-a)^{2}+(s-b)^{2}+(s-c)^{2}+s^{2}=a^{2}+b^{2}+c^{2}$

13) If $2s=a+b+c$, prove that $(s-a)^{3}+(s-b)^{3}+(s-c)^{3}+3abc=s^{3}$

14) If $2s=a+b+c$, prove that $16s(s-a)(s-b)(s-c)=2b^{2}c^{2}+2c^{2}a^{2}+2a^{2}b^{2}-a^{4}-b^{4}-c^{4}$

15) If   $2s=a+b+c$, then prove that  $2(s-a)(s-b)(s-c)+a(s-b)(s-c)+b(s-c)(s-a)+c(s-a)(s-b)=abc$

16) If $a+b+c=0$, then prove that $(2a-b)^{3}+(2b-c)^{3}+(2c-a)^{3}=3(2a-b)(2b-c)(2c-a)$

17) If $a+b+c=0$, then prove that $\frac{a^{2}}{2a^{2}+bc} + \frac{b^{2}}{2b^{2}+ca} + \frac{c^{2}}{2c^{2}+ab} =1$

18) Prove that $(x+y+z)^{3}+(x+y-z)^{3}+(x-y+z)^{3}+(x-y-z)^{3}=4x(x^{2}+3y^{2}+3z^{2})$

19) If $a+b+c=0$ prove that $(s+3a)^{3}-(s-3b)^{3}-(s-3c)^{3}-3(s-3a)(s-3b)(s-3c)=0$

20) If $X=b+c-2a$, $Y=c+a-2b$, $Z=a+b-2c$, find the value of $X^{2}+Y^{2}+Z^{2}-3XYZ$

21) Prove that $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=2(c-b)(c-a)+2(b-a)(b-c)+2(a-b)(a-c)$

22) Prove that $a^{2}(b^{3}-c^{3})+b^{2}(c^{3}-a^{3})+c^{2}(a^{3}-b^{3})=(a-b)(b-c)(c-a)(ab+bc+ca)=a^{2}(b-c)^{3}+b^{2}(c-a)^{3}+c^{2}(a-b)^{3} = -[a^{2}b^{2}(a-b)+b^{2}c^{2}(b-c)+c^{2}a^{2}(c-a)]$

23) if $(a+b)^{2}+(b+c)^{2}+(c+a)^{2}=4(ab+bc+cd)$, prove that $a=b=c=d$.

24) If $x=a+d$, $y=b+d$, $z=c+d$, prove that $x^{2}+y^{2}+z^{2}-yz-zx-xy=a^{2}+b^{2}+c^{2}-bc-ca-ab$

25) If $a+b+c=3$, prove that $\frac{1}{b^{2}+c^{2}-a^{2}}+ \frac{1}{c^{2}+a^{2}-b^{2}} + \frac{1}{a^{2}+b^{2}-c^{2}}=0$

26) If $a+b+c=0$, simplify: $\frac{b+c}{bc}(b^{2}+c^{2}-a^{2}) + \frac{c+a}{ca} (c^{2}+a^{2}-b^{2})+ \frac{a+b}{ab}(a^{2}+b^{2}-c^{2})$

27) Prove that the equation $(x-a)^{2}+(y-b)^{2}+(a^{2}+b^{2}-1)(x^{2}+y^{2}-1)=0$ is equivalent to the equation $(ax+by-1)^{2}+(bx-ay)^{2}=0$, hence show that the only possible values of x and y are: $\frac{a}{a^{2}+b^{2}}$, $\frac{b}{a^{2}+b^{2}}$

28) If $2(x^{2}+a^{2}-ax)(y^{2}+b^{2}-by)=x^{2}y^{2}+a^{2}b^{2}$, prove that $(x-a)^{2}(y-b)^{2}+(bx-ay)^{2}=0$ and therefore that $a=x$ and $y=b$ are the only possible solutions.

Good luck for the PreRMo August 2019 !!

Regards,

Nalin Pithwa