Question

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Question No. 55

Answer 1

Hey friend, Harish here.

Here is your answer:

55)

$2 \sqrt[3]{4} + 7 \sqrt[3]{32} - \sqrt[3]{500}$   -- (Given expression)

⇒ $2 \sqrt[3]{4} + 7 \sqrt[3]{4\times 8} - \sqrt[3]{4 \times 125}$

⇒ $2 \sqrt[3]{4} + 7 \sqrt[3]{4\times 2^{3}} - \sqrt[3]{4 \times 5^{3}}$

⇒ $2 \sqrt[3]{4} + (7 \times 2 \sqrt[3]{4}) - (5\sqrt[3]{4})$

⇒ $2 \sqrt[2]{4} + 14 \sqrt[3]{4} - 5 \sqrt[3]{4}$

⇒ $(2 + 14 - 5)( \sqrt[3]{4}) = 11 \sqrt[3]{4}$

Therefore the answer is (OPTION - C).

56)

$( \frac{a}{b})^{x-1} = ( \frac{b}{a})^{x-3}$

Now use the law of indices 1/x⁻ᵃ  = xᵃ.

Then,

⇒ $( \frac{a}{b})^{x-1} = ( \frac{a}{b})^{-(x-3)}$

Now as the bases are same we can compare the powers,

Then,

(x -1) = - (x -3)

⇒ (x - 1) = 3 - x

⇒ x + x = 3+ 1

⇒ 2x = 4.

⇒ x = 2.

Therefore the value of x is 2.
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Hope my answer is helpful to you.

Answer 2

The answer is given below :

$2 \sqrt[3]{4} + 7 \sqrt[3]{32} - \sqrt[3]{500} \\ \\ = 2( {4}^{ \frac{1}{3} } ) + 7( {32}^{ \frac{1}{3} } ) - ( {500}^{ \frac{1}{3} } ) \\ \\ =( 2 \times {( {2}^{2}) }^{ \frac{1}{3} }) +( 7 \times ( { {2}^{5} })^{ \frac{1}{3} } ) - ( { {5}^{3} \times {2}^{2} })^{ \frac{1}{3} } \\ \\ = (2 \times {2}^{ \frac{2}{3} } ) + (7 \times {2}^{ \frac{5}{3} } ) - (( { {5}^{3}) }^{ \frac{1}{3} } \times ( { {2}^{2} })^{ \frac{1}{3} } ) \\ \\ = (2 \times {2}^{ \frac{2}{3} } ) + (7 \times {2}^{(1 + \frac{2}{3} )} ) - (5 \times {2}^{ \frac{2}{3} } ) \\ \\ = (2 \times {2}^{ \frac{2}{3} } ) + (7 \times {2}^{1} \times {2}^{ \frac{2}{3} } ) - (5 \times {2}^{ \frac{2}{3} } ) \\ \\ = (2 \times {2}^{ \frac{2}{3} } ) + (14 \times {2}^{ \frac{2}{3} } ) - (5 \times {2}^{ \frac{2}{3} } ) \\ \\ = (2 + 14 - 5) \times {2}^{ \frac{2}{3} } \\ \\ = 11 \times {2}^{ \frac{2}{3} } \\ \\ = 11 \times( { {2}^{2} })^{ \frac{1}{3} } \\ \\ = 11 \times {4}^{ \frac{1}{3} } \\ \\ = 11 \times \sqrt[3]{4} \\ \\ = 11 \sqrt[3]{4}$

So, Option (C) is right.

Thank you for your question.