Question

If x² + 1/x² = 7 , find the value of x³ + 1/x³

Answer 1

Answer:

18

Step-by-step explanation:

Given Equation is x² + (1/x²) = 7.

It can be written as,

⇒ x² + (1/x²) = 9 - 2

⇒ x² + (1/x²) + 2 = 9

⇒ (x + 1/x)² = 9

⇒ x + 1/x = 3.

On cubing both sides, we get

⇒ (x + 1/x)³ = (3)³

⇒ x³ + 1/x³ + 3(x + 1/x) = 27

⇒ x³ + 1/x³ + 3(3) = 27

⇒ x³ + 1/x³ + 9 = 27

⇒ x³ + 1/x³ = 18.

Hope it helps!

Answer 2

Question :

If  $\sf \red{x^2 + \dfrac{1}{x^2}=7}$ , find the value of  $\sf \red{x^3 + \dfrac{1}{x^3}}$

Solution :

$\sf x^2+\dfrac{1}{x^2}=7$

$\longrightarrow\ \sf x^2+\dfrac{1}{x^2}=9-2$

$\longrightarrow\ \sf x^2+\dfrac{1}{x^2}+2=9$

$\longrightarrow\ \sf x^2+\dfrac{1}{x^2}+2.x.\dfrac{1}{x}=9$

$\bullet\ \; \sf \orange{(A+B)^2=A^2+B^2+2AB}$

$\longrightarrow\ \sf \left( x+ \dfrac{1}{x} \right)^2=9$

$\longrightarrow\ \sf \textsf{\textbf{x}}\ \textsf{\textbf{+}}\ \dfrac{\textsf{\textbf{1}}}{\textsf{\textbf{x}}}\ \textsf{\textbf{=}}\ \textsf{\textbf{3}}$  ∀ x ∈ N

Cubing on both sides ,

$\longrightarrow \sf \left( x+ \dfrac{1}{x} \right)^3 =3^3$

$\bullet\ \; \sf \blue{(A+B)^3=A^3+B^3+3AB(A+B)}$

$\longrightarrow \sf x^3+\dfrac{1}{x^3} + 3.x.\dfrac{1}{x} \left( \textsf{\textbf{x}}\ \textsf{\textbf{+}}\ \dfrac{\textsf{\textbf{1}}}{\textsf{\textbf{x}}} \right) = 27$

$\longrightarrow \sf x^3+\dfrac{1}{x^3} + 3(3)= 27$

$\longrightarrow \sf x^3+\dfrac{1}{x^3}= 27-9$

$\longrightarrow \sf \pink{\textsf{\textbf{x}}^{\textsf{\textbf{3}}}\ \textsf{\textbf{+}}\ \dfrac{\textsf{\textbf{1}}}{\textsf{\textbf{x}}^{\textsf{\textbf{3}}}}\ \textsf{\textbf{=}}\ \textsf{\textbf{18}}}$

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