Question 150

If $$x^6 - 512 y^6 = (x^2 + Ay^2) (x^4 - Bx^2 y^2 + Cy^4)$$, then what is the value of $$(A + B - C)$$?

Solution

$$x^6 - 512 y^6 = (x^2 + Ay^2) (x^4 - Bx^2 y^2 + Cy^4)$$

$$(x^2)^3 + (-8y^2)^3 = (x^2 + Ay^2) (x^4 - Bx^2 y^2 + Cy^4)$$

$$(\because a^3 + b^3 = (a + b)(a^2 - ab + b^2))$$

by comparison -

$$-8y^2 = Ay^2$$

A = -8

$$Bx^2y^2 = x^2 \times -8y^2$$

B = -8

$$(-8y^2)^2 = Cy^4$$

C = 64

The value of $$(A + B - C)$$ = -8 - 8 - 64 = -80

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