Surds and Indices Practice Test 2

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Question : 13 of 17

Marks: +1, -0

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

(x^b/x^{-c})^{b^2-bc+c^2}

(x^c/x^{-a})^{c^2-ca+a^2}

is equivalent to:

$x^{(a^2+b^2+c^2)}$
$x^{2(a^3+b^3+c^3)}$
$x^{2(a^3-b^3-c^3)}$
$1/x^{2(a^3+b^3+c^3)}$

Solution:

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

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

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

=x^{a^3+b^3}

Similarly,

(x^b/x^{-c})^{b^2-bc+c^2}

=x^{b^3+c^3}

and

(x^c/x^{-a})^{c^2-ca+a^2}

=x^{c^3+a^3}

∴

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

(x^b/x^{-c})^{b^2-bc+c^2}

(x^c/x^{-a})^{c^2-ca+a^2}

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

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