Question

if x2-3x-1=0 find the value of x-1/x and x3-1/x3

Answer 1

EXPLANATION.

⇒ x² - 3x - 1 = 0.

As we know that,

We can write equation as,

⇒ x² - 1 = 3x.

⇒ (x² - 1)/x = 3.

⇒ (x²/x) - (1/x) = 3.

⇒ (x - 1/x) = 3.

Now, cubing on both sides of the equation, we get.

⇒ (x - 1/x)³ = (3)³.

⇒ x³ - 3(x²)(1/x) + 3(x)(1/x²) - 1/x³ = 27.

⇒ x³ - 3x + 3/x - 1/x³ = 27.

⇒ x³ - 1/x³ - 3(x - 1/x) = 27.

Put the values of (x - 1/x) = 3 in the equation, we get.

⇒ x³ - 1/x³ - 3(3) = 27.

⇒ x³ - 1/x³ - 9 = 27.

⇒ x³ - 1/x³ = 27 + 9.

⇒ x³ - 1/x³ = 36.

Values of :

(1) = (x - 1/x) = 3.

(2) = (x³ - 1/x³) = 36.

Answer 2

Answer:

$\orange{ \boxed { \boxed{ \begin{array}{cc}\hookrightarrow \: \sf \: given \\ \\ \rm \: {x}^{2} - 3x - 1 = 0 \: \: \: - - -eqn. (1)\\ \\ \blue{ \underline{ \sf \: we \: have \: to \: find \: the \: value \: of : }} \\ \\ (a) \: \: \rm \: x - \frac{1}{x} \\ \\ \rm \: (b) \: \: \: {x}^{3} - \frac{1}{ {x}^{3} } \end{array}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \underline{ \sf \: solution}}$

$\pink{ \boxed{ \boxed{ \begin{array}{cc} (a) \\ \\ \sf \: from \: \: eqn.(1) = > \\ \\ \rm \: {x}^{2} - 3x - 1 = 0 \\ \\ \rm \implies \: {x}^{2} - 1 = 3x \\ \\ \rm \implies \: \frac{ {x}^{2} - 1}{x} = \frac{3x}{x} \\ \\ \rm \implies \: \frac{ {x}^{2} }{x} - \frac{1}{x} = 3 \\ \\ \rm \implies \:x - \frac{1}{x} = 3 \\ \\ \blue{ \boxed{ \rm \therefore \: x - \frac{1}{x} = 3}} \: \: \: - - -eqn. (2)\\ \\ \end{array}}}}$

$\orange{ \boxed{\boxed{\begin{array}{cc}(b) \\ \\ \rm \: we \: know \: that : \\ \\ \boxed{\sf \: {a}^{3} - {b}^{3} = ( {a - b)}^{3} + 3ab(a - b)} \\ \\ \bf \: now \\ \\ \small{\rm \: {x}^{3} - \frac{1}{ {x}^{3} } = ( {x - \frac{1}{x} })^{3} + 3 \times x \times \frac{1}{x} \times (x - \frac{1}{x} )} \\ \\ \rm = {3}^{3} + 3 \times 3 \: \: \red{\{ \sf \: from \: eqn.(2) \}} \\ \\ \rm = 27 + 9 \\ \\ = 36 \\ \\ \blue{ \boxed{ \rm \therefore \: {x}^{3} - \frac{1}{ {x}^{3} } = 36}} \\ \\ \end{array}}}}</p><p>$