Answer:
The value of a = 8.
The value of b = 2.
Step-by-step explanation:
Given: [tex]$ \sqrt[3]{x^{10}} $[/tex] where, x = -2
[tex]$ x^{-10} = (-2)^{-10} $[/tex]
Since, the power is even (10), the resultant will be a positive number.
So, we have:
[tex]$ \sqrt[3]{(-2)^{-10}} = \sqrt[3]{2^{10}} $[/tex]
Note that: [tex]$ \sqrt[a]{x} = x^{\frac{1}{a}} $[/tex]
Also, [tex]$ x^a . x^b = x^{a + b} $[/tex]
Therefore, we get:
[tex]$ \sqrt[3]{2^{10}} = 2^{\frac{10}{3}} $[/tex]
[tex]$ = 2^{\frac{9 + 1}{3}} $[/tex]
[tex]$ = 2^{\frac{9}{3}} . 2^{\frac{1}{3}} $[/tex]
[tex]$ = 2^3 . \sqrt[3]{2} $[/tex]
Hence, a = 8; b = 2
Hence, the answer.
