≤ft({x^a}{x^{-b}})^{a²-ab+b² ≤ft({x^b}{x^{-c}})^{b²-bc+c² ≤ft({x^c}{x^{-a}})^{c²-ca+a² is equivalent to:

Correct Answer is : x2(a3+b3+c3)x²(a³+b³+c³)}x2(a3+b3+c3)

Correct Answer is : x2(a3+b3+c3)x²(a³+b³+c³)}x2(a3+b3+c3)

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Surds and Indices Practice Test 2

Question : 13 of 17

Marks: +1, -0

\left(\frac{x^a}{x^{-b}}\right)^{a^2-ab+b^2}

\left(\frac{x^b}{x^{-c}}\right)^{b^2-bc+c^2}

\left(\frac{x^c}{x^{-a}}\right)^{c^2-ca+a^2}

is equivalent to:

Solution:

Concept:

The expression simplifies using the identity

a^3+b^3=(a+b)(a^2-ab+b^2)

and laws of indices.

Explanation:

First term:

\left(\frac{x^a}{x^{-b}}\right)^{a^2-ab+b^2} = \left(x^{a+b}\right)^{a^2-ab+b^2}

= x^{(a+b)(a^2-ab+b^2)} = x^{a^3+b^3}

Second term:

\left(\frac{x^b}{x^{-c}}\right)^{b^2-bc+c^2} = x^{b^3+c^3}

Third term:

\left(\frac{x^c}{x^{-a}}\right)^{c^2-ca+a^2} = x^{c^3+a^3}

Multiply all three terms:

x^{a^3+b^3} \cdot x^{b^3+c^3} \cdot x^{c^3+a^3}

= x^{(a^3+b^3)+(b^3+c^3)+(c^3+a^3)}

= x^{2(a^3+b^3+c^3)}

Answer:

x^{2(a^3+b^3+c^3)}

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