Question

If x+1/x=5 then find x3+1/x3?​

Answer 1

Answer:

$\large{\underline{\boxed{\sf {x}^{3} + \frac{1}{ {x}^{3} } = 110}}}$

Step-by-step explanation:

Given that-

$\implies \sf x + \frac{1}{x} = 5 \\ \\ \bf On \: cubing \: both \: sides : \\ \\ \implies \sf (x + \frac{1}{x} ) {}^{3} = (5) {}^{3} \\ \\ \implies \sf {x}^{3} + \frac{1}{ {x}^{3} } + 3 \: . \:x \: . \: \frac{1}{x} (x + \frac{1}{x} ) = 125 \\ \\ \implies \sf {x}^{3} + \frac{1}{ {x}^{3} } + 3(x + \frac{1}{x} ) = 125 \\ \\ \implies \sf {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 5 = 125 \\ \\ \implies \sf {x}^{3} + \frac{1}{ {x}^{3} } + 15 = 125 \\ \\ \implies \sf {x}^{3} + \frac{1}{ {x}^{3} } = 125 - 15 \\ \\ \boxed{ \boxed{ \bf \therefore \: {x}^{3} + \frac{1}{ {x}^{3} } = 110}}$

_______________________

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

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

Answer 2

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

$\mathfrak{\large{\underline{\underline{Explanation}}}}$

$\boxed{\bf{x^3 + \dfrac{1}{x^3} = 110}}$

Given :- $\sf{x + \dfrac{1}{x}=6}$

To find :- $\sf{x^3 + \dfrac{1}{x^3}}$

Solution :-

$\sf{x + \dfrac{1}{x} = 5}$

By cubing on both the sides

$\tt{{ \left( x + \dfrac{1}{x} \right) }^3 = 5^3}$

We know that, (a + b)³ = a³ + b³ + 3ab(a + b)

Here a = x, b = 1/x

By substituting the values in the identity we have

$\tt{x^3 + \dfrac{1}{x^3} + 3 \times x \times \dfrac{1}{x} \left( x + \dfrac{1}{x} \right) = 125}$

$\tt{x^3 + \dfrac{1}{x^3} + 3 \left( x + \dfrac{1}{x} \right) = 125}$

$\tt{x^3 + \dfrac{1}{x^3} + 3(5) = 125}$

[Since x + 1/x = 5]

$\tt{x^3 + \dfrac{1}{x^3} + 15 = 125}$

$\tt{x^3 + \dfrac{1}{x^3} = 125 - 15}$

$\tt{x^3 + \dfrac{1}{x^3} = 110}$

$\boxed{\bf{x^3 + \dfrac{1}{x^3} = 110}}$

$\mathfrak{\large{\underline{\underline{Identity\:Used:-}}}}$

(a + b)³ = a³ + b³ + 3ab(a + b)

$\mathfrak{\large{\underline{\underline{Some\:Important\:Identities:-}}}}$

[1] (a + b)² = a² + b² + 2ab

[2] (a - b)² = a² + b² - 2ab

[3] (a + b)(a - b) = a² - b²

[4] (x + a)(x + b) = x² + (a + b)x + ab