Question

If x ^2 + 1/4x ^ 2= 8 , find x^3 + 1/ 8 x ^3​

Answer 1

Answer:

x=3+

3

x

2

=(3+

8

)

2

=9+8+6

8

=17+6

8

x

2

+

x

2

1

=(3+

8

)

2

+

(3+

8

)

2

1

⇒17+6

18

+

17+6

8

1

⇒

17+6

8

(17+6

8

)

2

+1

×

(17−6

8

)

(17−6

8

)

⇒

289−288

(17+6

18

)

2

(17−6

8

)+(17−6

8

)

⇒(289−288)(17+6

8

)+17−6

8

⇒17+6

8

+17−6

8

=34

Answer 2

${x}^{2} + \frac{1}{ {4x}^{2} } = 8 \\ \\ {x}^{3} + \frac{1}{ {8x}^{3} } = \\ \\ {(x + \frac{1}{2x}})^{2} = {x}^{2} + \frac{1}{ {4x}^{2} } + 2 \times x \times \frac{1}{2x} \\ \\ {(x + \frac{1}{2x}})^{2} = 8 + 1 \\ \\ {(x + \frac{1}{2x}})^{2} = 9 \\ \\ {(x + \frac{1}{2x}}) = \sqrt{9} \\ \\ {(x + \frac{1}{2x}}) = 3.$

Case 1 :

$({x + \frac{1}{2x} })^{3} = {x}^{3} + \frac{1}{ {8x}^{3} } + 3 \times x \times \frac{1}{x} (x + \frac{1}{x} ) \\ \\ {3}^{3} = {x}^{3} + \frac{1}{ {8x}^{3} } + \frac{9}{2} \\ \\ 27 = {x}^{3} + \frac{1}{ {8x}^{3} } + \frac{9}{2} \\ \\ {x}^{3} + \frac{1}{ {8x}^{3} } = 27 - \frac{9}{2} \\ \\ {x}^{3} + \frac{1}{ {8x}^{3} } = \frac{54 - 9}{2} \\ \\ {x}^{3} + \frac{1}{ {8x}^{3} } = \frac{45}{2} \\ \\ {x}^{3} + \frac{1}{ {8x}^{3} } = 22 \times \frac{1}{2} .$

Case 2 :

${x}^{3} + \frac{1}{ {8x}^{3} } = - 27 + \frac{9}{2} \\ \\ {x}^{3} + \frac{1}{ {8x}^{3} } = \frac{ - 54 + 9}{2} \\ \\ {x}^{3} + \frac{1}{ {8x}^{3} } = \frac{ - 45}{2} \\ \\ {x}^{3} + \frac{1}{ {8x}^{3} } = - 22 \frac{1}{2} .$