Rewrite as 3c^4(c^3 -27) = 0
This is true if either c=0 or c^3 = 27
27 is the cube of 3, so c=3 is the other solution.
i really need help on these..i don't understand ....i need to find the real number solutions of the equations...
-2v^6=16
3s^4+15s^2-72=0
3c^7=81c^4
3q^4+3q^3=6q^2+6q
any help would be appreciate...thank you!!
2 answers
can you help me with the other three!!