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Question
By using identity evaluate the following:
73 – 103 + 33
Sum
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Solution
73 – 103 + 33
(a + b + c)(a2 + b2 + c2 – ab – bc – ca) = a3 + b3 + c3 – 3abc
If a + b + c = 0, then a3 + b3 + c3 = 3abc
∴ 7 – 10 + 3 = 0
⇒ 73 – 103 + 33 = 3 × 7 × –10 × 3
= 9 × – 70 = – 630
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Chapter 3: Algebra - Exercise 3.4 [Page 102]
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