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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