Question

X=1/4-x find the value of
X3+1/x3​

Answer 1

Answer:

$\large{\underline{\boxed{\sf {x}^{3} + \frac{1}{ {x}^{3} } = 52}}}$

Step-by-step explanation:

Given that-

$\sf x = \frac{1}{4 - x} \\ \\ \implies \sf \frac{1}{x} = 4 - x \\ \\ \implies \sf \frac{1}{x} + x = 4 \\ \\ \implies \sf x + \frac{1}{x} = 4 \\ \\ \bf on \: cubing \: both \: sides - \\ \\ \implies \tt (x + \frac{1}{x}) {}^{3} = (4) {}^{3} \\ \\ \implies \tt {x}^{3} + \frac{1}{ {x}^{3} } + 3 \: . \: x \: .\: \frac{1}{x} (x + \frac{1}{x} ) = 64 \\ \\ \implies \tt {x}^{3} + \frac{1}{ {x}^{3} } + 3(x + \frac{1}{x} ) = 64 \\ \\ \implies \tt {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 4 = 64 \\ \\ \implies \tt {x}^{3} + \frac{1}{ {x}^{3} } + 12 = 64 \\ \\ \implies \tt {x}^{3} + \frac{1}{ {x}^{3} } = 64 - 12 \\ \\ \implies \tt {x}^{3} + \frac{1}{ {x}^{3} } = 52$

Hence,

$\boxed{ \bf \therefore \: {x}^{3} + \frac{1}{ {x}^{3} } = 52}$