First, multiply both sides by 64 to remove the fraction. Remember, (1/X) x X = 1.

t^3 = 64t

then divide both sides by t

t^2 = 64

so, t = sq root of 64, = 8

t = -8, 0, and +8.

Given:

(1/64)t^3 = t

Subtract “t” from the right side:

(1/64)t^3 – t = 0

Now, factor out a “t” from both terms:

t * [(1/64)t^2 – 1] = 0

You now have:

t = 0 <--- First Factor (1/64)t^2 - 1 = 0 <-- Second Factor On the Second Factor, add 1 to both sides: (1/64)t^2 = 1 Now, multiply both sides by 64, which gives you: t^2 = 64 Taking the square root of both sides: t = ±8 If you graph this equation, you'll see the x-intercepts are: -8, 0 and +8. Good luck in your studies, ~ Mitch ~

t{(1/64)t^2-1} = 0

so t = 0 or t = 64^0.5, t = 8