If ( 1-x^6 + 1-y^6 =a (x^3-y^3 ) ), then (y^2 d y d x = )

Step-by-step solution

Given equation is ( aligned & 1-x^6 + 1-y^6 =a (x^3-y^3 ) & 1-x^6 + 1-y^6 x^3-y^3 =a aligned ) On differentiating both sides w.r.t., (x ), we get ( [ (x^3-y^3 ) ( -6 x^5 2 1-x^6 - 6 y^5 2 1-y^6 d y d x )- ( 1-x^6 + 1-y^6 ) (3 x^2-3 y^2 d y d x ) ] (x^3-y^3 )^2 =0 ) ( aligned & (y^2 ( 1-x^6 + 1-y^6 )- y^5 (x^3-y^3 ) 1-y^6 ) d y d x & =x^2 ( 1-x^6 + 1-y^6 )+ x^5 1-x^6 (x^3-y^3 ) & y^2 d y d x [ 1-x^6 + 1-y^6 + (1-y^6 )-y^3 x^3+y^6 1-y^6 ] & =x^2 [ (1-x^6 )+ 1-y^6 1-x^6 +x^6-x^3 y^3 1-x^6 ] & y^2 d y d x [ 1-x^6 1-y^6