Question

If 4x+ 1/5x = 2, then 125x^3+ 1/64x^3 is (a) 7 hold 1/4 (b) 6 hold 1/4 (c) 5 hold 1/4 (d) 4 hold 1/4

Answer 1

$Given \: 4x + \frac{1}{5x} = 2$

$Multiplying \: both \:sides\: of \: the$

$equation\: by \: \frac{5}{4} ,\:we \:get$

$\implies \frac{5}{4} \Big(4x + \frac{1}{5x} \Big)= 2 \times \frac{5}{4}$

$\implies \frac{5}{4} \times 4x +\frac{5}{4} \times \frac{1}{5x} \Big)= \frac{5}{2}$

$\implies 5x + \frac{1}{4x} = \frac{5}{2}\: --(1)$

/* By Algebraic Identity */

$\boxed {\pink { a^{3}+b^{3} = (a+b)^{3} - 3ab(a+b)}}$

$Now, \red{ Value \:of \: 125x^{3} + \frac{1}{64x^{3}} }$

$= (5x)^{3} + \Big(\frac{1}{4x}\Big)^{3}$

$= \Big( 5x + \frac{1}{4x}\Big)^{3} - 3\times 5x \times \frac{1}{4x} \Big( 5x + \frac{1}{4x}\Big)$

$= \Big( \frac{5}{2}\Big)^{3} - \frac{15}{4} \times \frac{5}{2}$

$= \frac{5}{2} \Big ( \big(\frac{5}{2}\big)^{2} - \frac{15}{4}\Big)$

$= \frac{5}{2} \Big( \frac{25}{4} - \frac{15}{4}\Big)$

$= \frac{5}{2}\Big( \frac{25-15}{4}\Big)$

$= \frac{5}{2} \times \frac{10}{4}$

$= \frac{25}{4}$

$\green {= 6\frac{1}{4}}$

Therefore.,

$Option \: \pink { ( b ) } \:is \: correct.$

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Answer 2

Answer:

option b 6 whole 1/4 is your correct answer.

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