Question

Find the value of x³+1/x³,if x²+1/x²=14

Answer 1

Answer:

$\large{\underline{\boxed{\bf x^3 + \dfrac{1}{x^3} = 52}}}$

Step-by-step explanation:

Given that -

$\tt {x}^{2} + \dfrac{1}{ {x}^{2} } = 14$

Adding 2 both sides -

$\tt {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 14 + 2 \\ \\ \tt {x}^{2} + \frac{1}{ {x}^{2} } + 2 \: .\: x \: . \: \frac{1}{x} = 16 \\ \\ \tt (x + \frac{1}{x} ) {}^{2} = 16 \\ \\ \tt x+ \frac{1}{x} = \sqrt{16} \\ \\ \implies \tt x + \frac{1}{x} = 4$

On cubing both sides -

$\tt (x + \frac{1}{x} ) {}^{3} = (4)^{3} \\ \\ \tt x^{3} + \frac{1}{x^{3}} + 3 \: . \: x \:. \: \frac{1}{x} (x + \frac{1}{x} ) = 64 \\ \\ \tt x^{3} + \frac{1}{x^{3}} + 3(4) = 64 \\ \\ \tt x^{3} + \frac{1}{x^{3}} + 12 = 64 \\ \\ \tt x^{3} + \frac{1}{x^{3}} = 64 - 12 \\ \\ \red{\boxed{\tt x^{3} + \frac{1}{x^{3}} = 52}}$

Hence, the answer is 52.

$\rule{300}{2}$

$\large{\underline{\underline{\mathfrak{\heartsuit \: Algebraic \: Identities : \: \heartsuit}}}}$

• (a - b)³ = a³ - b³ - 3ab (a - b)
• (a + b)³ = a³ + b³ + 3ab (a + b)
• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• (x + a)(x + b) = x² + x(a + b) + ab

Answer 2

$\huge\textbf{Answer:-}$

According to the given question:-

$x {}^{2} + \frac{1}{x {}^{2} } = 14$

We need to add the number(2) to both sides:-

So,

$= > x {}^{2} + \frac{1}{x {}^{2} } + 2 = 14 + 2$

$= > x {}^{2} + \frac{1}{x {}^{2} } + 2(x) \frac{1}{x} = 16$

$= > (x + \frac{1}{x} ) {}^{2} = 16$

$= > x + \frac{1}{x} = 4$

Cubing both the sides (3)

$= > (3 + \frac{1}{x} ) {}^{3} = (4) {}^{3}$

$= > x {}^{3} + \frac{1}{x {}^{3} } = 3(x) \frac{1}{x} (x + \frac{1}{x} = 64 )$

$= > x {}^{3} + \frac{1}{x {}^{3} } + (3)....(4 = 64)$

$= > x {}^{3} + \frac{1}{x {}^{3} } + 12 = 64$

$= > x {}^{3} + \frac{1}{x {}^{3} } = 64 - 12$

$= > x {}^{3} + \frac{1}{x {}^{3} } = 52$

Therefore, 52 is the answer:-