CBSE Class 10 Answered

Let's consider a³ + b³, for example,

(a+b)³ = a³+b³ + 3ab(a+b)

a³+b³= (a+b)³- 3ab(a+b)

(a³+b³)(a+b) = [(a+b)³- 3ab(a+b)]/(a+b) = (a+b)²- 3ab

Similarly for in general using the binomial expansion,

(a+b)ⁿ = aⁿ + na^n-1b + n(n-1)a^n-2b²/2 ...............................n(n-1)a²b^n-2/2 + nab^n-1 + bⁿ

But the second and second last term,

na^n-1b + nab^n-1= n(aⁿb/a + bⁿa/b) = n(aⁿb² + bⁿa²)/ab = nab(a^n-2+b^n-2)

Now since n is odd, n-2 will be odd, and successive we can reduced the power until we get (a+b). This can be done with all the terms and we can divide by (a+b)

Using the same arguments we can prove the same, for aⁿ - bⁿ.

Regards,

Team,

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