simplify a(a2 + a + 1) + 5 and find its values for (i) a = 0, (ii) a = 0, and (iii) a = -1.
Subtract: 3a(a + b + c) – 2b(a – b + c) from 4c(–a + b + c)
∵   3a(a + b + c) – 2b(a – b + c)
                    = (3a x a + 3a x b + 3a x c) -(2b x a + 2b x (-b) + 2b x c)
                    = (3a2 + 3ab + 3ac) - (2ab - 2b2 + 2bc)
                    = 3a2 + 3ab + 3ac- 2ab - 2b2 + 2bc
                    = 3a2 + ab + 3ac + 2b2 - - 2ab
and 4c(-a + b + c) = 4c x (-a) + 4c x b +4c x c
                         = -4ac + 4bc + 4c2Â
∴   -4ac + 4bc + 4c2
       3ac  - 2bc + 3a2 + ab + 2b2Â
       (-)     (+)    (-)  (-)  (-)
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(-4 - 3)ac + ( 4 + 2)bc + 4c2 -3a2 -ab -2b2
                            = -7ac + 6bc + 4c2 -3a2
                            = -3a2--2b2+4c2-ab+6 bc - 7ac