If a+b+c=0, then find the value of frac{a^{4} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $a+b+c=0.$Squaring both sides,we get $(a+b+c)^2 = 0^2.$$a^2+b^2+c^2+2(ab+bc+ca) = 0 \Rightarrow a^2+b^2+c^2 = -2(ab+bc+ca).$Squaring again,$

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given $a+b+c=0.$Squaring both sides,we get $(a+b+c)^2 = 0^2.$$a^2+b^2+c^2+2(ab+bc+ca) = 0 \Rightarrow a^2+b^2+c^2 = -2(ab+bc+ca).$Squaring again,$(a^2+b^2+c^2)^2 = [-2(ab+bc+ca)]^2.$$a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2) = 4(a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)).$Since $a+b+c=0,$ the term $2abc(a+b+c) = 0.$So,$a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2) = 4(a^2b^2+b^2c^2+c^2a^2).$$a^4+b^4+c^4 = 4(a^2b^2+b^2c^2+c^2a^2) - 2(a^2b^2+b^2c^2+c^2a^2) = 2(a^2b^2+b^2c^2+c^2a^2).$Therefore,$\frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2} = 2.$

If $a+b+c=0,$ then find the value of $\frac{a^{4}+b^{4}+c^{4}}{b^{2} c^{2}+c^{2} a^{2}+a^{2} b^{2}}.$

Similar Questions

$I. x^{2}-20 x+91=0$
$II. y^{2}-32 y+247=0$

If the equation $\frac{1}{x} + \frac{1}{x - 1} + \frac{1}{x - 2} = 3x^3$ has $k$ real roots,then $k$ is equal to -

The expression $x^{4}+7 x^{2}+16$ can be factored as:

Solve the given two equations and select the correct option.
$I.$ $12x^2 - 47x + 40 = 0$
$II.$ $4y^2 + 3y - 10 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module