Use the rules of exponents to simplify the expression $3\left( 2 x ^ { - 5 } y ^ { 5 } \right) ^ { - 3 } \left( 2 x ^ { - 5 } y ^ { 5 } \right) ^ { 3 }$ using only positive exponents. Assume that no variable is zero.

A) $64y30x30\frac { 64 y ^ { 30 } } { x ^ { 30 } }$
B) 1
C) $64x30y30\frac { 64 x ^ { 30 } } { y ^ { 30 } }$
D) $−4y30x30\frac { - 4 y ^ { 30 } } { x ^ { 30 } }$
E) 64

Question

Use the rules of exponents to simplify the expression

3\left( 2 x ^ { - 5 } y ^ { 5 } \right) ^ { - 3 } \left( 2 x ^ { - 5 } y ^ { 5 } \right) ^ { 3 }

using only positive exponents. Assume that no variable is zero.

A)

64y30x30\frac { 64 y ^ { 30 } } { x ^ { 30 } }

B) 1
C)

64x30y30\frac { 64 x ^ { 30 } } { y ^ { 30 } }

D)

−4y30x30\frac { - 4 y ^ { 30 } } { x ^ { 30 } }

E) 64

Dimas Fatur · Accepted Answer

Final Answer : B, Explanation :   Applying the rules of exponents, we get ( 2 x − 5 y 5 ) − 3 ( 2 x − 5 y 5 ) 3 = 2 − 3 x 15 y − 15 ⋅ 2 3 x − 15 y 15 = 2 − 3 ⋅ 2 3 ⋅ x 15 ⋅ x − 15 ⋅ y − 15 ⋅ y 15 1 = 1.  \left( 2 x ^ { - 5 } y ^ { 5 } ight) ^ { - 3 } \left( 2 x ^ { - 5 } y ^ { 5 } ight) ^ { 3 }  & = 2 ^ { -3 } x ^ { 15 } y ^ { -15 } \cdot 2 ^ { 3 } x ^ { -15 } y ^ { 15 } \ & = \frac { 2 ^ { -3 } \cdot 2 ^ { 3 } \cdot x ^ { 15 } \cdot x ^ { -15 } \cdot y ^ { -15 } \cdot y ^ { 15 } } { 1 } \ & = 1.  ( 2 x − 5 y 5 ) − 3 ( 2 x − 5 y 5 ) 3 ​ = 2 − 3 x 15 y − 15 ⋅ 2 3 x − 15 y 15 = 1 2 − 3 ⋅ 2 3 ⋅ x 15 ⋅ x − 15 ⋅ y − 15 ⋅ y 15 ​ = 1. ​

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