Question

If x+1/x=3, then find the value of x³+1/ x³​

Answer 1

S O L U T I O N :

$\underline{\bf{Given\::}}$

If x + 1/x = 3..........(1)

$\underline{\bf{Explanation\::}}$

$\longrightarrow\tt{x + \dfrac{1}{x} = 3}$

Cubing on both sides :

$\longrightarrow\tt{\bigg(x + \dfrac{1}{x}\bigg)^{3} = (3)^{3}}$

$\longrightarrow\tt{(x)^{3} +\bigg(\dfrac{1}{3} \bigg)^{3} + 3 \times x \times 1/x (x + 1/x) = 27}$

$\longrightarrow\tt{(x)^{3} +\bigg(\dfrac{1}{3} \bigg)^{3} + 3 \times \cancel{x} \times 1/\cancel{x} (3) = 27\:\:\:[from(1)]}$

$\longrightarrow\tt{(x)^{3} +\bigg(\dfrac{1}{3} \bigg)^{3} + 3(3) = 27}$

$\longrightarrow\tt{(x)^{3} +\bigg(\dfrac{1}{3} \bigg)^{3} + 9 = 27}$

$\longrightarrow\tt{(x)^{3} +\bigg(\dfrac{1}{3} \bigg)^{3} =27-9}$

$\longrightarrow\tt{(x)^{3} +\bigg(\dfrac{1}{3} \bigg)^{3} =18}$

Thus,

The value of x³ + 1/x³ will be 18 .

Answer 2

$\huge{ \underline{ \tt{Given \ratio}}}$

$\bf \: x + \frac{1}{x } = 3$

Now on cubing both the sides; we get;

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

$\tt \{using \: (a + b) {}^{3} = {a}^{3} + {b}^{3} + 3a {}^{2} b + 3ab {}^{2} \}$

$\bf \to(x {)}^{3} + ( \frac{1}{x} {)}^{3} + 3(x {)}^{2} \frac{1}{x} + 3x( \frac{1}{x} {)}^{2} = 27$

$\bf \to {x}^{3} + \frac{1}{ {x}^{3} } + 3. { \cancel{x}}^{2} . \frac{1}{ \cancel{x}} + 3. \cancel{x}. \frac{1}{ { \cancel{x}}^{2} } = 27$

$\bf \to {x}^{3} + \frac{1}{ {x}^{3} } + 3x + 3 \frac{1}{x} = 27$

Now on taking out commons; we get;

$\bf \to {x}^{3} + \frac{1}{ {x}^{3} } + 3(x + \frac{1}{x} ) = 27$

As we know,

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

$\bf \to {x}^{3} + \frac{1}{ {x}^{3} } + 3(3) = 27$

$\bf \to {x}^{3} + \frac{1}{ {x}^{3} } + 9 = 27$

$\bf \to {x}^{3} + \frac{1}{ {x}^{3} } = 27 - 9$

$\bf \to {x}^{3} + \frac{1}{ {x}^{3} } = \red{18}$

Therefore,

$\tt the \: value \: of \: \boxed{ \bf{x}^{3} + \frac{1}{ {x}^{3} } = 18}$