Question

$\sf\pink{If \: x+ \dfrac{1}{x} =3} \: \underline{\red{Calculate}} \: \: \: x²+ \dfrac{1}{x²} \: \: , \: \: x³+ \dfrac{1}{x³} \\ \tt{and \: x⁴+ \dfrac{1}{x⁴}}$

$\textit {Quality}\:\textsl {answer }\:\textsc {required}$​

Answer 1

Identity used -

$\to \bold \pink{(a+b)² = a²+b² +2ab}\\ \\ \to \bold \orange{(a +b)³ = a³+b³ +3ab(a+b)}$

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Solution

~Here the concept of algebric identity is used. We should know few identities and in order to find the values.

$\mathtt{Its \: given \: that \:\color{skyblue} x+ \frac{1}{X} = 3}$

Now square both the sides

$\sf{\implies (x+\frac{1}{x} )²= 3²} \\ \\ \sf{\implies x²+\frac{1}{x²}+ 2×x+\frac{1}{x} = 9} \\ \\ \sf{ \implies x²+\frac{1}{x²}= 9-2}\\ \\ \sf{\implies \color{red} x²+\frac{1}{x²}=7 }$

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Cube both the sides

$\sf{\implies (x+\frac{1}{x} )³= 3³} \\\\ \sf{ \implies x³+\frac{1}{x³} +3×x×\frac{1}{x}(x+\frac{1}{x} )=27 }\\\\ \sf{\implies x³+\frac{1}{x³}+3(x+\frac{1}{x}) = 27} \\ \\ \sf{ \implies {x}^{3} + \frac{1}{ {x}^{3 } } + 3 \times 3 = 27} \\ \\ \sf{ \implies \color{magenta}{x}^{3} + \frac{1}{ {x}^{3 } } = 18}$

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We already know that

$\sf{\bull \color{red} x²+\frac{1}{x²}=7 }$

Square both the sides

$\sf{\implies (x²+\frac{1}{x²})^{2} =7^{2} } \\ \\ \sf{ \implies {x}^{4} + \frac{1}{ {x}^{4} } + 2 {x}^{2} \times \frac{1}{x ^{2}} = 49 } \\ \\ \sf{ \implies {x}^{4} + \frac{1}{ {x}^{4} } = 49 - 2} \\ \\ \sf{ \implies \color{orange}{x}^{4} + \frac{1}{ {x}^{4} } = 47 }$

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More identities!!

♦(x + y)² = x² + 2xy + y²

♦(x – y)² = x² – 2xy + y²

♦x²– y² = (x + y) (x – y)

♦(x + a) (x + b) = x2 + (a + b)x + ab.

♦(x + y + z)² = x² + y² + z²+ 2xy + 2yz + 2zx.

♦(x + y)³ = x³ + y³ + 3xy(x + y)

♦(x – y)³ = x³ – y³ – 3xy(x – y)

♦x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y²+ z² – xy – yz – zx)

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