Inactive Tutor answered 03/13/13
You have two more complex solutions.
Use the sum of two cubes' formula,
t^3 + 216 = t^3 + 6^3 = (t+6)(t^2-6t+36) = 0
t + 6 = 0 => t = -6
or
t^2-6t+36 = 0
=> t^2-6t = -36
=> (t-3)^2 = 9-36 = -27
t = 3 +/- 3sqrt(3) i
Courtnee J.
asked 03/13/13How do I solve t3+216? Thank you for your help!
Inactive Tutor answered 03/13/13
You have two more complex solutions.
Use the sum of two cubes' formula,
t^3 + 216 = t^3 + 6^3 = (t+6)(t^2-6t+36) = 0
t + 6 = 0 => t = -6
or
t^2-6t+36 = 0
=> t^2-6t = -36
=> (t-3)^2 = 9-36 = -27
t = 3 +/- 3sqrt(3) i
Inactive Tutor answered 03/13/13
t3 + 216 = 0
Isolate the variable.
t3 = -126
Now take the cube root of each side, since the cube root of a number is the inverse of a number cubed.
√(t3) = √(-216) = -6
t = -6
Since (-6)(-6)(-6) = (-6)^3 = -216
Inactive Tutor
Yes, the second factorization from the sum of two cubes. Robert has correct my error above. As he pointed out, there should be two more complex solutions!
03/13/13
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Courtnee J.
Thanks Anthony! Does that mean it would be (t-216)(t+1)(t-1) or (t-6)(t^2+6t+36)?03/13/13