Given, function y = x3 + 21
Diff w.r.t. t,
dy/dx = 3x2 (dx/dt)…..(i)
According to question
dx/dt = 3(dx/dt) …..(ii)
From equation (i) and (ii)
3x2 = 3
⇒ x = ± 1
In equation y = x3 + 21
Putting x = 1
y = 13 + 21
= 1 + 21 =22
Putting x = -1
y = (- 1)3 + 21 = -1 + 21 = 20
So, x = ± 1 and y = 22, 20