Math, asked by vinitajain62, 11 months ago

if \: x ^{2} + \frac{1}{4x ^{2} } = 8 \: then \: find \: x ^{3} + \frac{1}{8x ^{3} }
mathematics: expansion.
don't spam or joke.corono ki kasam.​
slide to view the question.

Answers

Answered by RvChaudharY50
53

||✪✪ 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 + + 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 = + + 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 + 1/8x³ is ±22.5.

꧁________________________꧂

Answered by EliteSoul
72

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} }

Similar questions