If the roots of the equation $(c^2-ab)x^2-2(a^2-bc)x+b^2-ac=0$ are equal, prove that either $a=0$ or $a^3+b^3+c^3=3abc$.

Given:

Given quadratic equation is $(c^2-ab)x^2-2(a^2-bc)x+b^2-ac=0$. The roots of the given quadratic equation are equal.

To do:

We have to prove that either $a=0$ or $a^3+b^3+c^3=3abc$.

Solution:

Comparing the given quadratic equation with the standard form of the quadratic equation $ax^2+bx+c=0$, we get,

$a=(c^2-ab), b=-2(a^2-bc)$ and $c=(b^2-ac)$.

The discriminant of the standard form of the quadratic equation $ax^2+bx+c=0$ is $D=b^2-4ac$.

$D=[-2(a^2-bc)]^2-4(c^2-ab)(b^2-ac)$

$D=4(a^4-2a^2bc+b^2c^2)-4(b^2c^2-ac^3-ab^3+a^2bc)$

$D=4a^4-8a^2bc+4b^2c^2-4b^2c^2+4ac^3+4ab^3-4a^2bc$

$D=4a^4+4ac^3+4ab^3-12a^2bc$

$D=4a(a^3+c^3+b^3-3abc)$

The given quadratic equation has equal roots if $D=0$.

This implies,

$4a(a^3+c^3+b^3-3abc)=0$

$4a=0$ or $a^3+c^3+b^3-3abc=0$

$a=0$ or $a^3+c^3+b^3=3abc$

Hence proved.

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Updated on: 10-Oct-2022

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