Mathematics
When a = 3, b = 0 and c = 4, find the value of:
(i) ab + 2bc + 3ca + 4abc
(ii) a3 + b3 + c3 - 3abc
Answer
Given, a = 3, b = 0 and c = 4.
(i) Substituting a = 3, b = 0 and c = 4 in ab + 2bc + 3ca + 4abc, we get:
ab + 2bc + 3ca + 4abc
= (3)(0) + 2(0)(4) + 3(4)(3) + 4(3)(0)(4)
= 0 + 0 + 36 + 0
= 36.
∴ ab + 2bc + 3ca + 4abc = 36.
(ii) Substituting a = 3, b = 0 and c = 4 in a3 + b3 + c3 - 3abc, we get:
a3 + b3 + c3 - 3abc
= (3)3 + (0)3 + (4)3 - 3(3)(0)(4)
= 27 + 0 + 64 - 0
= 91.
∴ a3 + b3 + c3 - 3abc = 91.
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