Simplify: \(\dfrac{\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}}{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}\)
\((a+b)(b+c)(c+a)\)
Both numerator and denominator are sums of three cubes, which is the cue for the conditional identity: if \(x+y+z=0\), then \(x^{3}+y^{3}+z^{3}=3xyz\).
For the numerator take \(x=a^{2}-b^{2},\ y=b^{2}-c^{2},\ z=c^{2}-a^{2}\). Their sum telescopes to \((a^{2}-b^{2})+(b^{2}-c^{2})+(c^{2}-a^{2})=0\), so the numerator becomes \(3(a^{2}-b^{2})(b^{2}-c^{2})(c^{2}-a^{2})\).
For the denominator take \(x=a-b,\ y=b-c,\ z=c-a\). Again the sum is \((a-b)+(b-c)+(c-a)=0\), so the denominator becomes \(3(a-b)(b-c)(c-a)\).
Now factor each numerator bracket as a difference of squares: \(a^{2}-b^{2}=(a-b)(a+b)\), \(b^{2}-c^{2}=(b-c)(b+c)\), and \(c^{2}-a^{2}=(c-a)(c+a)\).
The whole expression is now \(\frac{3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)}{3(a-b)(b-c)(c-a)}\).
Cancel the factors common to top and bottom — the \(3\), and each of \((a-b),(b-c),(c-a)\) — leaving only the sum-factors.
The simplified expression is \((a+b)(b+c)(c+a)\).
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Important Questions from Identities
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(999) 2- 2 2=
