SOLUTION: 8(c3 - c) = c (6c + 1) or 8(c^3-c) = c (6c+1)

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Question 80737: 8(c3 - c) = c (6c + 1)
or
8(c^3-c) = c (6c+1)
Answer by jim_thompson5910(21685) About Me  (Show Source):
You can put this solution on YOUR website!
8%28c%5E3-c%29+=+c+%286c%2B1%29

8c%5E3-8c+=+6c%5E2%2Bc Distribute

8c%5E3-6c%5E2-8c-c+=+0 Get all terms to one side

8c%5E3-6c%5E2-9c+=+0 Combine like terms

c%288c%5E2-6c-9%29+=+0 Factor out a c

Now factor 8c%5E2-6c-9
note: this solver uses x instead of c
Error, solver not defined for name 'quadratic_factoring'.
Error occurred executing solver 'quadratic_factoring' .



So we have

c%284c%2B3%29%282c-3%29+=+0
which gives us

c=0 or 4c%2B3=0 or 2c-3=0 Set each individual factor equal to zero

c=0 or 4c=-3 or 2c=3 Now solve for c in each case

So our answer is

c=0 or c=-3%2F4 or c=3%2F2