If the sum of the cubes of the roots of the equation $x^3-a x^2+b x-c=0$ is zero, then $a^3+3 c=$

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Aicharya · Accepted Answer

Given: $x^3-a x^2+b x-c=0 ...(i)$ Let $\alpha, \beta \& \gamma$ are the roots equation (i), we get $\begin{aligned} & \alpha+\beta+\gamma=a \ & \alpha \beta+\beta \gamma+\gamma \alpha=b \ & \alpha \beta \gamma=c \end{aligned}$ Also, given: $\alpha^3+\beta^3+\gamma^3=0$ $\begin{aligned} & \Rightarrow(\alpha+\beta+\gamma)\left(\alpha^2+\beta^2+\gamma^2-\alpha \beta-\beta \gamma-\gamma \alphaight)+3 \alpha \beta \gamma=0 \ & \Rightarrow a\left((\alpha+\beta+\gamma)^2-3(\alpha \beta+\beta \gamma+\gamma \alpha)ight)+3 c=0 \ & \Rightarrow a\left[a^2-3 bight]+3 c=0 \ & \Rightarrow a^3-3 a b+3 c=0 \Rightarrow a^3+3 c=3 a b \end{aligned}$

If the sum of the cubes of the roots of the equation $x^3-a x^2+b x-c=0$ is zero, then $a^3+3 c=$

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