Question

If x^2 + + 1/x^2 = 7, Find x^3 + 1/x^3​

Answer 1

Answer:

Given:

${x}^{2} + \frac{1}{x {}^{2} } = 7$

To find:

${x}^{3} + \frac{1}{x {}^{3} }$

Solution:

$\Longrightarrow$ ${x}^{2} + \frac{1}{x {}^{2} } = {x}^{2} + \frac{1}{x {}^{2} } + 2 \times x \times \frac{1}{x} = 7 + 2$

$\Longrightarrow$ x + 1/x = √9 = 3

Therefore,

$(x + \frac{1}{x} ) {}^{3} = (x + \frac{1}{x} )( {x}^{2} + \frac{1}{ {x}^{2} } - 1) \\ \\ = 3 \times (7 - 1) = 3 \times 6 = 18$

Some algebraic identities :

→ (a + b)^2 = a^2 + 2ab + b^2

→ (a – b)^2 = a^2 – 2ab + b^2

→ a^2 – b^2 = (a + b) (a – b)

→ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

→ (a + b – c)^2 = a^2 + b^2 + c^2 + 2ab – 2bc – 2ca

→ (a – b – c)^2 = a^2 + b^2 + c^2 – 2ab + 2bc – 2ca

→ (a + b)^3 = a^3 + b^3 + 3ab(a + b)

→ (a – b)^3 = a^3 – b^3 – 3ab(a – b)

→ (a^3 + b^3) = (a + b) (a^2 – ab + b^2)

→ (a^3 – b^3) = (a – b) (a^2 + ab + b^2)