Question

please solve it
$\huge if \: x + \frac{1}{x} = 2$
$\huge \: then \: {x}^{3 } + \frac{1}{ {x}^{3} }$

​

Answer 1

Step-by-step explanation:

Formula:

(a+b)^3 = a^3+b^3+3ab(a+b)

(x+1/X)^3 = 2

cube on both sides

(x+1/X)^3 = 8

x^3 + (1/X)^3 + 3×X×1/X (X+1/X) = 8

x^3 + 1/x^3 + 3(2) = 8

x^3 + 1/x^3 + 6 = 8

x^3 + 1/x^3 = 8-6

x^3 + 1/x^3 = 2

Answer 2

$\huge{\mathfrak{\red{\dag{\pink{\underline{\underline{\orange{Solution}}}}}}}}$

a³+b³ = (a+b)(a²+ab+b²)

a = x ; b = 1/x

x³ + (1/x)³ = (x+1/x)[x²+x*1/x+(1/x²)]

= 2 * [x²+(1/x)²]

= 2 * (x+1/x)²

= 2 * 2²

= 2 * 4 = 8