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Algebra and Applications Test 3

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Question : 2 of 25

Marks: +1, -0

If

x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}

y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}

, then the value of

x^{3} + y^{3}

is:

730
650
970
950

Solution:

Concept:

Use

xy=1

and the sum-of-cubes identity to simplify the calculation.

Explanation:

Rationalize

x

:

x=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{(\sqrt{3}-\sqrt{2})^2}{3-2}=5-2\sqrt{6}

.

Similarly,

y=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=5+2\sqrt{6}

.

So,

x+y=(5-2\sqrt{6})+(5+2\sqrt{6})=10

.

Also,

xy=(5-2\sqrt{6})(5+2\sqrt{6})=25-24=1

.

Use the identity:

x^3+y^3=(x+y)^3-3xy(x+y)

Substitute the values:

x^3+y^3=10^3-3(1)(10)

=1000-30=970

Answer:

970

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