Question

Solve this Question please​

Answer 1

Given expression,

$\sf \sqrt[3]{ {x}^{2} } + 3 \sqrt[3]{x} = 4$

Assume y = x⅓, substituting value of 'y' in the above expression, we obtain :

$\longrightarrow \sf {y}^{2} + 3y - 4 = 0 \\ \\ \longrightarrow \sf {y}^{2} + 4y - y - 4 = 0 \\ \\ \longrightarrow \sf(y + 4)(y - 1) = 0 \\ \\ \longrightarrow \sf ( \sqrt[3]{x} + 4)( \sqrt[3]{x} - 1) = 0 \\ \\ \longrightarrow \sf \: \sqrt[3]{x} = - 4 \: or \: 1 \\ \\ \longrightarrow \boxed{ \boxed{\sf x = - 64 \: or \: 1}}$