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Question : 4 of 160
Marks:
+1,
-0
Solution:
We have,
||‌C2⟶C2−C1‌ and ‌C3⟶C3−C1‌=|| 1 | 0 | 0 |
| a2 | b2−a2 | c2−a2 |
| a3 | b3−a3 | c3−a3 |
|‌=(b−a)(c−a)‌|| 1 | 0 | 0 |
| a2 | (b+a) | (c+a) |
| a3 | (b2+a2+ab) | (c2+a2+ac) |
|‌=(b−a)(c−a)‌||‌=[(b−a)(c−a)[bc2+a2b+abc+ac2+a3..‌+a2c−b2c−a3−a2b−a2c−abc−ab2]‌=(b−a)(c−a)[bc2+ac2−b2c−ab2]‌=(b−a)(c−a)[bc(c−b)‌+a(c−b)(c+b)]‌=(a−b)(b−c)(c−a)(ab+bc+ca)
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