Question

If x+1/x=4,then find a3+ 1/a3​

Answer 1

Answer:

$x {}^{3} + \frac{1}{x {}^{3} } = 52$

Step-by-step explanation:

$\boxed{given}$

$x + \frac{1}{x} = 4$

$\boxed{Tofind}$

$x {}^{3} + \frac{1}{x {}^{3} }$

$\boxed{answer}$

given

$x {}^{} + \frac{1}{x {}^{} } = 4$

cubing both sides

we get

$(x + \frac{1}{x} ) {}^{3} = (4) {}^{3}$

using identity

$\green{(a+b)^3=a^3+b^3+3xy(x+y)}$

$x {}^{3} + \frac{1}{x {}^{3} } + 3(x)( \frac{1}{x} )(x + \frac{1}{x} ) = 64$

$x + \frac{1}{x} = 4...given$

so we get

$x {}^{3} + \frac{1}{x {}^{3} } + 3(4) = 64$

$x {}^{3} + \frac{1}{x {}^{3} } + 12 = 64$

$x {}^{3} + \frac{1}{x {}^{3} } = 64 - 12$

$x {}^{3} + \frac{1}{x {}^{3} } = 64 - 12$

$x {}^{3} + \frac{1}{x {}^{3} } = 52$