Question

x^2+1/4x^2=8 find x^3+1/8x^3​

Answer 1

Given:

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

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

Solution:

∴ $x^{2} +\dfrac{1}{4x^{2}} =8$

⇒ $x^{2} +(\dfrac{1}{2x})^2 =8$

Using the algebraic identity:

$a^{2}$ + $b^2$ = $(a+b)^{2}$ - 2ab

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

⇒ $(x+\dfrac{1}{2x})^2$ + 2 = 8

⇒ $(x+\dfrac{1}{2x})^2$ = 8 + 2 = 10

⇒ $x+\dfrac{1}{2x}$ = $\sqrt{10}$

∴ $x^{3} +\dfrac{1}{8x^{3}}$

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

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

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

= 10$\sqrt{10}$ - 3$\sqrt{10}$

= 7$\sqrt{10}$

∴ $x^{3} +\dfrac{1}{8x^{3}}$ = 7$\sqrt{10}$

Thus, the value of $x^{3} +\dfrac{1}{8x^{3}}$ is "equal to 7$\sqrt{10}$".

Answer 2

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