Question

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

Answer 1

Given, equation is:

$(c^2–ab)x^2–2(a^2–bc)x+(b^2–ac)=0Toprove:a=0or{a}^3+b^3+c^3=3abc$

Proof: From the given equation, we have

$A=(c^2–ab)B=–2(a^2–bc)C=(b^2–ac)$

It is being given that the equation has real and equal roots

$\therefore D=0\Rightarrow B^2–4AC=0$

On substituting respective values of a, b and c in the above equation, we get

$[–2(a^2–bc){]}^2–4(c^2–ab\left)\right(b^2–ac)=04(a^2–bc{)}^2–4(c^2{b}^2–a{c}^3–a{b}^3+a^{2}bc)=04(a^4+b^2{c}^2–2a^{2}bc)–4(c^2{b}^2–a{c}^3–a{b}^3+a^{2}bc)=0\Rightarrow a^4+b^2{c}^2–2a^{2}bc–b^2{c}^2+a{c}^3+a{b}^3–a^{2}bc=0\Rightarrow a^4+a{b}^3+a{c}^3–3a^{2}bc=0\Rightarrow a[a^3+b^3+c^3–3abc]=0\Rightarrow a=0or{a}^3+b^3+c^3=3abc$