To find : Value of (√3+√2)6-(√3-√2)6
Formula used: (i)
nCr = \(\frac{n!}{(n-r)!(r)!}\)
(ii) (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + … +nCn-1abn-1 + nCnbn
= 2[(6)a5b + (20)a3b3 + (6)ab5]
⇒ (a+b)6 - (a-b)6 = 2[(6)a5b + (20)a3b3 + (6)ab5]
Putting the value of a = √3 and b = √2 in the above equation
(√3 + √2)6 - (√3 - √2)6
⇒ 2[(6)(√3)5(√2)+(20)(√3)3(√2)3 + (6)(√3)(√2)5]
⇒ 2[54(√6) + 120(√6) + 24 (√6)]
⇒ 396√6