Question 70

If a+b=1, then $$a^{4}+b^{4}-a^{3}-b^{3}-2a^{2}b^{2}+ab$$

Solution

$$a^{4}+b^{4}-a^{3}-b^{3}-2a^{2}b^{2}+ab$$

= $$a^{4}-2a^{2}b^{2}+b^{4}-a^{3}-b^{3}+ab$$

=($$a^{2}-b^{2})^2$$-$$((a+b)^3-3ab(a+b))$$+ab

=$$((a+b)(a-b))^2$$-$$[(1)^3-3ab(1)]$$+ab

=$$((1)(a-b))^2$$-(1-3ab)+ab

=$$(a-b)^2$$-1+3ab+ab

=$$(a-b)^2$$+4ab-1

=$$(a+b)^2$$-1

=$$(1)^2$$-1

=0

Therefore, Option D is the right choice.

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