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

 

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