Question

if x+1/x =4 find x square + 1 by x square and x cube +1 by x cube​

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

→ (x + 1/x) = 4

squaring both sides,

→ (x + 1/x)² = 4²

using (a + b)² = a² + b² + 2ab in LHS,

→ x² + 1/x² + 2 * x * 1/x = 16

→ (x² + 1/x²) + 2 = 16

→ (x² + 1/x²) = 16 - 2

→ (x² + 1/x²) = 14 (Ans.)

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Now,

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

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

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

Putting both values in RHS now,

→ (x³ + 1/x³) = 4 * (14 - 1)

→ (x³ + 1/x³) = 4 * 13

→ (x³ + 1/x³) = 52 (Ans.)

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Answer 2

${ \bold{ \red{ \underline{</strong><strong>GIVEN</strong><strong>-</strong><strong>}}}}$

${ \dagger{ \tt{ \purple{ \: \: \: x + \frac{1}{x} = 4}}}}$

${ \bold{ \red{ \underline{</strong><strong>TO</strong><strong>\: </strong><strong>FIN</strong><strong>D</strong><strong>-</strong><strong>}}}}$

${ \dagger{ \tt{ \purple{ \: \: \: {x}^{2} + \frac{1}{ {x}^{2} } \: \: and \: \: {x}^{3} + \frac{1}{ {x}^{3}}}}} }$

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${ \huge{ \rm{ \underline { \underline{ \blue{ </strong><strong>Solution-</strong><strong>}}}}}}$

1⃣. x^2 + 1/x^2

• we know that....

${ \boxed { \boxed{ \sf{ \green{ \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \: }}}}}$

using this algebraic identity for finding x^2 + 1/x^2

${ \rightarrow{ \pink{ \sf{ {(x + \frac{1}{x} )}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2 . x . \frac{1}{x}}}}}$

• putting the value of x + 1/x in the formula...

${ \implies{ \sf{ \pink{ {4}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2}}}}$

${ \implies{ \sf{ \pink{16 = {x}^{2} + \frac{1}{ {x}^{2} } + 2}}}}$

${ \implies{ \sf{ \pink { {x}^{2} + \frac{1}{ {x}^{2} } = 16 - 2}}}}$

${ \implies{ \underline{ \boxed{ \blue{ \sf{ {x}^{2} + \frac{1}{ {x}^{2} } = 14}}}}}}$

2⃣. x^3 + 1/x^3

• using the algebraic identity given below for finding x^3 + 1/x^3

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

similarly,

${ \rightarrow{ \pink{ \sf{ {x}^{3} + \frac{1}{ {x}^{3} } = (x + \frac{1}{x} )( {x}^{2} + \frac{1}{ {x}^{2} } - x. \frac{1}{x})}}} }$

• putting the values of x + 1/x and x^2 + 1/x^2. in the formula...

${ \implies{ \pink{ \sf{ {x}^{3} + \frac{1}{ {x}^{3} } = (4)(14 - 1)}}}}$

${ \implies{ \pink{ \sf{ {x}^{3} + \frac{1}{ {x}^{3} } = 4 \times 13}}}}$

${ \implies{ \underline{ \boxed{ \blue{ \sf{ {x}^{3} + \frac{1}{ {x}^{3} } = 52}}}}}}$