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Question : 17 of 120
Marks:
+1,
-0
Solution:
Take
(a+b)4=‌4C0a4+‌4C1a3b+‌4C2a2b2+‌4C3ab3+‌4C4b4
=‌4C0a4+‌4C1a3b+‌4C2a2b2+‌4C1ab3+‌4C0b4
(∵‌nCr=‌nCn−r) Similarly,
(a−b)4=a4−4a3b+6a2b2−4ab3+b4....(ii)
On subtracting Eq. (ii) from Eq. (i), we get
(a+b)4−(a−b)4=8a3b+8ab3 =8ab(a2+b2) Now, putting
a=√3 and
b=√2 (√3+√2)4−(√3−√2)4=8√3√2[(√3)2+(√2)2]
=8√6(3+2)=8√6×5=40√6
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