Question

$if \: x ^{2} + \frac{1}{4x ^{2} } = 8 \: then \: find \: x ^{3} + \frac{1}{8x ^{3} }$
mathematics: expansion.
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Answer 1

||✪✪ QUESTION ✪✪||

if x² + 1/4x² = 8 , find x³ + 1/8x³ = ?

|| ✰✰ ANSWER ✰✰ ||

→ x² + (1/4x²) = 8

Adding 1 both sides we get ,,

→ x² + (1/4x²) + 1 = 8 +1

Now, LHS , can be written as ,

→ (x)² + (1/2x)² + 2 * x * (1/2x) = 9

Comparing it with a² + b² + 2 * a * b = (a + b)² , we get,

→ ( x + 1/2x)² = 9

Square root both sides we get,

→ (x + 1/2x) = ±3

That Means,

→ (x + 1/2x) = 3 ---------- Equation (1)

→ (x + 1/2x) = (-3) ---------- Equation (2)

____________________________

Now, Taking Equation (1) value and cubing both sides we get,

→ (x + 1/2x)³ = 3³

Using (a+b)³ = a³ + b³ + 3ab(a+b) now, we get ,

→ x³ + 1/8x³ + 3*x*1/2x(x+1/2x) = 27

→ x³ + 1/8x³ + 3 *x *1/2x * 3 = 27

→ x³ + 1/8x³ + 9/2 = 27

→ x³ + 1/8x³ = 27 - (9/2)

→ x³ + 1/8x³ = (54-9)/2

→ x³ + 1/8x³ = 45/2

→ x³ + 1/8x³ = 22.5 (Ans).

___________________________

Now, Taking Equation (2) value and cubing both sides we get,

→ (x + 1/2x)³ = (-3)³

Using (a+b)³ = a³ + b³ + 3ab(a+b) now, we get ,

→ x³ + 1/8x³ + 3*x*1/2x(x+1/2x) = (-27)

→ x³ + 1/8x³ + 3 *x *1/2x * (-3) = (-27)

→ x³ + 1/8x³ - 9/2 = (-27)

→ x³ + 1/8x³ = (-27) + 9/2

→ x³ + 1/8x³ = (-54+9)/2

→ x³ + 1/8x³ = (-45)/2

→ x³ + 1/8x³ = (-22.5) (Ans).

Hence, The value of x³ + 1/8x³ is ±22.5.

꧁________________________꧂

Answer 2

Answer:

$\star\: \bf{x^3 + \dfrac{1}{8x^3}}$ = $\sf\green{\dfrac{45}{2}}$

Given:-

• $\rm x^2 + \dfrac{1}{4x^2} = 8$

To find:-

• $\rm x^3 + \dfrac{1}{8x^3} = ?$

Solution:-

First find the value of $\bold{x + \dfrac{1}{2x}}$

$\rule{100}{1}$

We know that,

$\dag \: \: \sf\blue{a^2 + b^2 = {(a + b)}^{2} - 2ab}$

$\Rightarrow\sf x^2 + \dfrac{1}{4x^2} = 8 \\\\\Rightarrow\sf (x + \dfrac{1}{2x})^2 - 2\times x \times \dfrac{1}{2x} = 8 \\\\\Rightarrow\sf (x + \dfrac{1}{2x})^2 - 1 = 8 \\\\\Rightarrow\sf (x + \dfrac{1}{2x})^2 = 8 + 1 \\\\\Rightarrow\sf (x + \dfrac{1}{2x}) = \sqrt{9} \\\\\Rightarrow{\boxed{\sf\green{x + \dfrac{1}{2x} = 3 }}}$

$\rule{200}{1}$

We know that,

$\dag\: \: \sf\blue{a^3 + b^3 = (a + b)^3 - 3ab(a + b) }$

$\Rightarrow\sf x^3 + \dfrac{1}{8x^3} = {(x + \dfrac{1}{2x})}^{3} - 3\times x \times \dfrac{1}{2x}(x + \dfrac{1}{2x})$

$\sf \: \: \: \: \: = {3}^{3} - \dfrac{3}{2}(3)$

$\sf \: \: \: \: \: = 27 - \dfrac{9}{2}$

$\sf \: \: \: \: \: = \dfrac{54 - 9}{2}$

$\sf\blue{\: \: \: \: \: = \dfrac{45}{2}}$

$\therefore\bold{Required \: value = \dfrac{45}{2} }$