Question

can any one solve it..​

Answer 1

Given:

$\bold{a^{2} + b^{2} + c^{2} = 4 abc}$

Question: Find the value of ${ \left( 4^{\frac{1}{bc}} \right) }^{a} \cdot { \left( 4^{\frac{1}{ac}} \right) }^{b} \cdot { \left( 4^{\frac{1}{ab}} \right) }^{c}$

SOLUTION

$</p><p>\rightarrow \quad { \left( 4^{\frac{1}{bc}} \right) }^{a} \cdot { \left( 4^{\frac{1}{ac}} \right) }^{b} \cdot { \left( 4^{\frac{1}{ab}} \right) }^{c}</p><p>\\ \\</p><p>\rightarrow \quad 4^{\frac{1}{bc} \times a} \cdot 4^{\frac{1}{ac} \times b} \cdot 4^{\frac{1}{ab} \times c} \qquad \left( \because (a^{m})^{n} = a^{mn} \right)</p><p>\\ \\</p><p>\rightarrow \quad 4^{\frac{a}{bc}} \cdot 4^{\frac{b}{ac}} \cdot 4^{\frac{c}{ab}} </p><p>\\ \\</p><p>\rightarrow \quad 4^{\frac{a}{bc} + \frac{b}{ac} + \frac{c}{ab}} \qquad \left( \because a^{m} \cdot a^{n} = a^{m+n} \right) </p><p>\\ \\</p><p>\rightarrow \quad 4^{\frac{a^{2} + b^{2} + c^{2}}{abc}}</p><p>\\ \\</p><p>\rightarrow \quad 4^{\frac{4\cancel{abc}}{\cancel{abc}}} \qquad \left( \because a^{2} + b^{2} + c^{2} = 4abc \quad \mathtt{given} \right) </p><p>\\ \\ </p><p>\rightarrow \quad 4^{4} \\ \\ \rightarrow \quad \bold{256}</p><p>$

Answer: Option D

Answer 2

Answer:

Option D

That is 256...