Show that the cubes of the roots of $x^{3} + a x^{2} + b x + a b = 0$ are given by the equation $x^{3} + a^{3} x^{2} + b^{3} x + a^{3} b^{3} = 0$ .

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Solution

Given equation, $x^{3} + a x^{2} + b x + a b = 0$ ....(1)

Roots of the required equation are cubes of the roots of given equation.

Therefore, $y = x^{3} \Rightarrow x = \sqrt[3]{y}$

Substituting value of $x$ in (1), we get

$(\sqrt[3]{y})^{3} + a (\sqrt[3]{y})^{2} + b (\sqrt[3]{y}) + a b = 0$

$\Rightarrow a \sqrt[3]{y^{2}} + b \sqrt[3]{y} = - (y + a b)$ ....(2)

$\Rightarrow a^{3} y^{2} + b^{3} y + 3 a b y (a \sqrt[3]{y^{2}} + b \sqrt[3]{y}) = - (a^{3} b^{2} + 3 a^{2} b^{2} y + 3 a b y^{2} + y^{3}) [Taking cubes on both sides]$

$\Rightarrow a^{3} y^{2} + b^{3} y + 3 a b y (- y - a b) = - (a^{3} b^{2} + 3 a^{2} b^{2} y + 3 a b y^{2} + y^{3}) [Using (2)]$

$\Rightarrow y^{3} + a^{3} y^{2} + b^{3} y + a^{3} b^{3} = 0$

Hence, proved.

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