(x + y) ^ 2 (X-Y) + (x + y) (Y-X) ^ 2 to extract the common factor

(x + y) ^ 2 (X-Y) + (x + y) (Y-X) ^ 2 to extract the common factor

(x+y)^2(x-y)+(x+y)(y-x)^2
=(x+y)(x-y)(x+y+x-y)
=2x(x+y)(x-y)

(M + n) ^ 2 - (X-Y) ^ 2
(2a+3b)^2-(a-2b)^2
4a^2n+4-9

A (X-Y) (x + y) - B (Y-X) (x + y) - C (X-Y) ^ 2 (y + x) to extract the common factor

Original formula = (x ^ 2-y ^ 2) a - (x ^ 2-y ^ 2) B - (x ^ 2-y ^ 2) (X-Y) C
=(x^2-y^2)(a-b-cx-cy)
=(x+y)(x-y)(a-b-cx-cy)

If the relation between the velocity V of a particle and the distance s is v = 1 + S ^ 2, then the expression of the tangential acceleration in S is

v=1+s²
a=dv/dt=(dv/ds)(ds/dt)
a=v(dv/ds)=v[d(1+s²)/ds]=2sv=2s(1+s²)