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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