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Question 88
If $$2\sqrt{2}x^3-3\sqrt{3}y^3=\left(\sqrt{2}x-\sqrt{3}y\right)\left(Ax^2+By^2+Cxy\right)$$, then the value of $$A^2 + B^2 - C^2$$ is:
Solution
$$2\sqrt{2}x^3-3\sqrt{3}y^3=\left(\sqrt{2}x-\sqrt{3}y\right)\left(Ax^2+By^2+Cxy\right)$$
($$\because$$ $$a^3 - b^3 = (a - b)(a^2 +Â ab + b^2)$$)
On compression,
A = $$(\sqrt{2})^2$$ = 2
A = $$(-\sqrt{3})^2$$ = 3
C = $$\sqrt{2}\sqrt{3} = \sqrt{6}$$
Now,
$$A^2 + B^2 - C^2$$
=Â $$2^2 + 3^2 - (\sqrt{6})^2$$
= 4 + 9 - 6
= 7
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