Evaluate $$(\sqrt{3}+\sqrt{2})^6- (\sqrt{3}-\sqrt{2})^6$$

Asked by Pragya Singh | 1 year ago |  114

Solution :-

Using binomial theorem the expression (a + b)6 and (a – b)6, can be expanded

(a + b)6 = 6C0 a6 + 6C1 a5 b + 6C2 a4 b2 + 6C3 a3 b3 + 6C4 a2 b4 + 6C5 a b6C6 b6

(a – b)6C0 a6 – 6C1 a5 b + 6C2 a4 b2 – 6C3 a3 b3 + 6C4 a2 b4 – 6C5 a b6C6 b6

Now (a + b)6 – (a – b)6 =6C0 a6 + 6C1 a5 b + 6C2 a4 b2 + 6C3 a3 b3 + 6C4 a2 b4 + 6C5 a b6C6 b6 – [6C0 a6 – 6C1 a5 b + 6C2 a4 b2 – 6C3 a3 b3 + 6C4 a2 b4 – 6C5 a b6C6 b6]

Now by substituting a = $$\sqrt{3}$$ and b =$$\sqrt{2}$$ we get

$$(\sqrt{3} + \sqrt{2})^6 -(\sqrt{3} - \sqrt{2})^6$$

$$2 [6 (\sqrt{3})^5 (\sqrt{2}) + 20 (\sqrt{3})^3 (\sqrt{2})^3 + 6 (\sqrt{3}) (\sqrt{2})^5]$$

$$2 [54(\sqrt{6}) + 120 (\sqrt{6}) + 24 \sqrt{6}]$$

$$2 (\sqrt{6}) (198)$$

$$396 \sqrt{6}$$

Answered by Abhisek | 1 year ago

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