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Question : 77 of 86

Marks: +1, -0

Solve for

x

x^{a^{3}} \cdot x^{b^{3}} \cdot x^{3ab(a+b)} = (2^{5})^{25}

a+b=5

2
3
1
0

Solution:

Concept:

Combine exponents using law

a^m \cdot a^n = a^{m+n}

and simplify using identity

a^3+b^3=(a+b)^3-3ab(a+b)

.

Explanation:

Left side:

x^{a^{3}} \cdot x^{b^{3}} \cdot x^{3ab(a+b)} = x^{a^3+b^3+3ab(a+b)}

.

Since

a^3+b^3 = (a+b)^3-3ab(a+b)

, exponent becomes

(a+b)^3-3ab(a+b)+3ab(a+b) = (a+b)^3

.

Given

a+b=5

, exponent =

5^3 = 125

. So left side =

x^{125}

.

Right side:

(2^5)^{25} = 2^{5 \times 25} = 2^{125}

.

Thus

x^{125} = 2^{125}

implies

x=2

(take 125th root).

Answer:

x = 2

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