If x=1+tt3,y=32t2+2t, then x(dydx)3−dydx is equal to
given,
x=1+tt3,y=32t2+2t
x=1t3+1t2,y=32t2+2t
dxdt=−3t4−2t3,dydt=−3t3−2t2
∴dydx=(dydt)(dxdt)=t
x(dydx)3−dydx=xt3−t
=1(∵x=1+tt3)