Question 830735
if a+b+c=4, then
a³+b³+c³-12c²+48c-64 is equal to
(A)3abc-12ab
(B)3abc+12ab
(C)abc+12ab
(D)3abc
please show how to solve this question

<pre>

          a+b+c = 4
            a+b = 4-c

Cube both sides:

         (a+b)³ = (4-c)³
a³+3a²b+3ab²+b³ = 64-48c+12c²-c³

We want to find a³+b³+c³-12c²+48c-64,
so we isolate that expression on the left

a³+b³+c³-12c²+48c-64 = -3a²b-3ab² = 

-3ab(b+a) = -3ab(a+b) = -3ab(4-c) = 

-12ab+3abc = 3abc-12ab

So the correct choice is (a). 

Edwin</pre>