Question

26. If x2 + 1/4x2= 8, find x3 +1/8x3​

Answer 1

Answer:

Given:

x^{2} +\dfrac{1}{4x^{2}} =8x

2

+

4x

2

1

=8

We have to find the value of x^{3} +\dfrac{1}{8x^{3}}x

3

+

8x

3

1

= ?

Solution:

∴ x^{2} +\dfrac{1}{4x^{2}} =8x

2

+

4x

2

1

=8

⇒ x^{2} +(\dfrac{1}{2x})^2 =8x

2

+(

2x

1

)

2

=8

Using the algebraic identity:

a^{2}a

2

+ b^2b

2

= (a+b)^{2}(a+b)

2

- 2ab

(x+\dfrac{1}{2x})^2+2x.\dfrac{1}{2x}=8(x+

2x

1

)

2

+2x.

2x

1

=8

⇒ (x+\dfrac{1}{2x})^2(x+

2x

1

)

2

+ 2 = 8

⇒ (x+\dfrac{1}{2x})^2(x+

2x

1

)

2

= 8 + 2 = 10

⇒ x+\dfrac{1}{2x}x+

2x

1

= \sqrt{10}

10

∴ x^{3} +\dfrac{1}{8x^{3}}x

3

+

8x

3

1

= (x+\dfrac{1}{2x})^3-3x.\dfrac{1}{2x}(x+\dfrac{1}{2x})(x+

2x

1

)

3

−3x.

2x

1

(x+

2x

1

)

= (x+\dfrac{1}{2x})^3-3(x+\dfrac{1}{2x})(x+

2x

1

)

3

−3(x+

2x

1

)

= (\sqrt{10} )^3-3\sqrt{10}(

10

)

3

−3

10

= 10\sqrt{10}

10

- 3\sqrt{10}

10

= 7\sqrt{10}

10

∴ x^{3} +\dfrac{1}{8x^{3}}x

3

+

8x

3

1

= 7\sqrt{10}

10

Thus, the value of x^{3} +\dfrac{1}{8x^{3}}x

3

+

8x

3

1

is "equal to 7\sqrt{10}

10

".

Explanation:

hope that helps you

plz Mark brainliest

Answer 2

What do you wrote there?