Question

if x^2+1/x^2=18, find the value of x^3-1/x^3​

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

→ x² + 1/x² = 18

subtracting 2 from both sides,

→ x² + 1/x² - 2 = 18 - 2

→ x² + 1/x² - 2 * x * 1/x = 16

comparing the LHS with a² + b² - 2ab = (a - b)² we get,

→ (x - 1/x)² = 16

Square root both sides now , we get ,

→ (x - 1/x) = ±4

____________

Now , we have :-

• (x² + 1/x²) = 18
• (x * 1/x) = 1
• (x - 1/x) = ±4

using the formula a³ - b³ = (a - b)(a² + b² + ab) Now, we get,

→ (x³ - 1/x³) = (x - 1/x)(x² + 1/x² + 1)

→ (x³ - 1/x³) = ±4(18 +1)

→ (x³ - 1/x³) = ±4 * 19

→ (x³ - 1/x³) = ±76 (Ans.)

________________________

Answer 2

${ \bold{ \huge{ \underline{ \underline{ \red{Solution:-}}}}}}$

${ \bold{ \purple{GIVEN- }}}$

${ \star{ \bold{ \: \: \: \: {x} ^{2} + \frac{1}{ {x}^{2} } = 18}}}$

${ \bold{ \purple{TO \: FIND- }}}$

${ \star{ \bold{ \: \: \: {x}^{3} - \frac{1}{ {x}^{3}}}} }$

▪ using the algebraic identity....

${ \boxed{ \bold{ \pink{ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab}}}}$

${ \boxed{ \bold{ \pink{ {a}^{2} + {b}^{2} = {(a - b)}^{2} + 2ab}}}}$

${ \bold{ \rightarrow{ {x}^{2} + \frac{1}{ {x}^{2} } = 18}}}$

${ \bold{ \implies{ {(x - \frac{1}{x} )}^{2} + 2x \times \frac{1}{x}} = 18}}$

${ \bold{ \implies{ {(x - \frac{1}{x} )}^{2} + 2 = 18}}}$

${ \bold{ \implies{ {(x - \frac{1}{x} )}^{2} = 18 - 2 = 16}}}$

${ \bold{ \implies{ \red{(x - \frac{1}{x} ) = \sqrt{16} = 4}}}}$

now, we have to find

${ \bold{ {x}^{3} - \frac{1}{ {x}^{3}}} }$

${ \boxed{ \pink{ \bold{ {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + {b}^{2} + ab}}}}$

using the above identity....

${ \bold{ \rightarrow{ {x}^{3} - \frac{1}{ {x}^{3} } = (x - \frac{1}{x} )( {x}^{2} + \frac{1}{ {x}^{2} } + x \times \frac{1}{x})}}}$

${ \bold{putting \: the \: values \: of}}$

${ \bold{x - \frac{1}{x} \: \: and \: \: {x}^{2} + \frac{1}{ {x}^{2} } }}$

${ \bold{ \implies{ {x}^{3} - \frac{1}{ {x}^{3} } = (4)(18 + 1)}}}$

${ \bold{ \implies{ {x}^{3} - \frac{1}{ {x}^{3} } = 4 \times 19}}}$

${ \boxed{ \boxed{ \bold{ \red{ {x}^{3} - \frac{1}{ {x}^{3} } = 76}}}}}$