Question

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

Answer 1

Answer:

$x + \frac{1}{x} = 3 \\ \\ squaring \: both \: side \: \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 9 \\ \\ \\ { \large{ \boxed{ \pink{{x}^{2} + \frac{1}{ {x}^{2} } = 7}}}}$

$x + \frac{1}{x} = 3 \\ \\ cubing \: both \: side \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times x \times \frac{1}{x} (x + \frac{1}{x} ) = 27 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 3 = 27 \\ \\ \\ { \large{ \boxed{ \pink{ {x}^{3} + \frac{1}{ {x}^{3} } = 18}}}}$

${x}^{2} + \frac{1}{ {x}^{2} } = 7 \\ \\ squaring \: both \: sides \\ \\ {x}^{4} + \frac{1}{ {x}^{4} } + 2 = 49 \\ \\{ \large{ \boxed{ \pink{ {x}^{4} + \frac{1}{ {x}^{4} } = 47}}}}$

Answer 2

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