Question

If x+1/2x=2,then find the value of 8x^3+ 1/x^3​

Answer 1

Question :

If x+(1/2x)=2,then find the value of 8x^3+ 1/x^3

Answer :

The value of 8x^3 + (1/x^3) is 40.

Given :

x+(1/2x)=2

To find :

The value of 8x^3+ 1/x^3

Solution :

From equation x+(1/2x)=2

Multiplying both sides by 2 , we get :

⇒ 2x + (2/2x) = 4

⇒ 2x + (1/x) = 4

Cubing both sides, we get :

⇒ {2x + (1/x)}^3 = (4)^3

We know, (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

⇒ 8x^3 + (1/x^3) + 3 × 2x × (1/x){2x + (1/x)} = 64

⇒ 8x^3 + (1/x^3) + 6 × 4 = 64

⇒ 8x^3 + (1/x^3) = 64 - 24

⇒ 8x^3 + (1/x^3) = 40

∴ The value of 8x^3 + (1/x^3) is 40.

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Answer 2

Answer:

The value of  $8x^3 + \frac{1}{x^3}$ = 40

Step-by-step explanation:

Given,

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

To find,

The value of $8x^3 + \frac{1}{x^3}$

Solution:

Recall the formula

(a+b) ³ = a³ + b³+3ab(a+b)

Given condition is $x+\frac{1}{2x} = 2$

Cubing on both sides, we get

$(x+\frac{1}{2x})^3 = 2^3= 8$

Applying the identity (a+b) ³ = a³ + b³+3ab(a+b) we get

$x^3+\frac{1}{(2x)^3} + 3 x\frac{1}{2x}(x+\frac{1}{2x}) = 8$

$x^3+\frac{1}{8x^3} + \frac{3}{2}(x+\frac{1}{2x}) = 8$

Substituting the value of  $x+\frac{1}{2x} = 2$ from the given condition we get,

$x^3+\frac{1}{8x^3} + \frac{3}{2}(2) = 8$

$x^3+\frac{1}{8x^3} + 3 = 8$

$x^3+\frac{1}{8x^3} = 8 - 3 =5$

Taking $\frac{1}{8}$ outside we get

$\frac{1}{8}(8x^3 + \frac{1}{x^3} ) = 5$

$8x^3 + \frac{1}{x^3} = 5X8 = 40$

∴ $8x^3 + \frac{1}{x^3}$ = 40

The value of  $8x^3 + \frac{1}{x^3}$ = 40

#SPJ2