Question

cube root x² +2x⅓-16x^-⅔-32/x by x⅙+4x^-⅙+4/ root ​

Answer 1

$\huge\mathscr\red{Hello \ Dude}$

We know that,

$\frac{1}{ {a}^{m} } = {a}^{ - m} ====== (1)$

$\frac{1}{32} = \frac{1}{ {2}^{5} } \\ = > {2}^{ - 5} .............using(1) \\ = > \frac{1}{32}= ( {2}^{ - 5} )= = = = = = = (2)$

$( \sqrt[3]{{4}})^{(2x + \frac{1}{2})} = {4}^{ \frac{1}{3} ( \frac{4x + 1}{2} ) } \\ = > {2}^{ 2 \times ( \frac{4x + 1}{6 } ) } = {2}^{( \frac{4x + 1}{3}) } = = = = = (3)$

according to questions

$( \sqrt[3]{{4}})^{(2x + \frac{1}{2})} = \frac{1}{32} \\ \\ USING --(2) AND--(3) \\ \\ = > {2}^{( \frac{4x + 1}{3}) } = {2}^{ - 5} \\ \\ = > \frac{4x + 1}{3} = - 5 \\ = > 4x + 1 = - 15 \\ = > 4x = - 15 - 1 \\ = > 4x = - 16 \\ \\ = > x = \frac{ - 16}{4} \\ = > x = - 4$

${ \ hope \ it's \ helps \ you}$

Answer 2

$\huge\bold{Answer:-}$

We know that,

we know that:-

$\frac{1}{ {a}^{m} } = {a}^{ - m} ====== (1)$

$\frac{1}{32} = \frac{1}{ {2}^{5} } \\ = > {2}^{ - 5} .............using(1) \\ = > \frac{1}{32}= ( {2}^{ - 5} )= = = = = = = (2)$

$( \sqrt[3]{{4}})^{(2x + \frac{1}{2})} = {4}^{ \frac{1}{3} ( \frac{4x + 1}{2} ) } \\ = > {2}^{ 2 \times ( \frac{4x + 1}{6 } ) } = {2}^{( \frac{4x + 1}{3}) } = = = = = (3)$

according to questions

$( \sqrt[3]{{4}})^{(2x + \frac{1}{2})} = \frac{1}{32} \\ \\ USING --(2) AND--(3) \\ \\ = > {2}^{( \frac{4x + 1}{3}) } = {2}^{ - 5} \\ \\ = > \frac{4x + 1}{3} = - 5 \\ = > 4x + 1 = - 15 \\ = > 4x = - 15 - 1 \\ = > 4x = - 16 \\ \\ = > x = \frac{ - 16}{4} \\ = > x = - 4$

${ \ hope \ it's \ helps \ you}$