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Question

If a=2,b=3,c=4, find the value of:

(a) 3ab+2c
(b) a2b2+c2
(c) ab3abc2ac
(d) a2b+bc3c32abc

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Solution

(a)3ab+2c=3×23+2×4=63+8=11
(b)a2b2+c2=2232+42=49+16=11
(c)ab3abc2ac=2×33×3×42×2×4=67216=82
(d)a2b+bc3c32abc=22×3+3×43×432×2×3×4 =12+1219248=216

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