is (x^3 + 2a^3 - 3xa^2) divisible by (x - a)^2 ?

=( x- a} ( x^2 + ax -2a^2 }

= {x-a) ( x^2 + 2ax - ax - 2a^2 }

= (x-a) { x(x+ 2a) - a( x+ 2a) }

= (x-a) ( x-a) ( x+2a)

= {x-a }^2 ( x+2a )

so it is divisible by ( x-a)^2

P(x) = x^3 + 2a^3 - 3xa^2

P(a) = a^3 + 2a^3 - 3a^3 = 0 then x- a is factor

P ' (x) = 3x^2 - 3a^2 = 3(x-a)(x+a) then x-a is double root (zeros)

P(x) is dividing by (x-a)^2

yes it is .

explanation---------------------

x^3 - 3xa^2 + 2a^3

put x= a

so a^3 - 3a^3 +2a^3 =0

so (x-a) one factor

x^3- 3xa^2 + 2a^3= ( x- a} ( x^2 + ax -2a^2 }

= {x-a) ( x^2 + 2ax - ax - 2a^2 }

= (x-a) { x(x+ 2a) - a( x+ 2a) }

= (x-a) ( x-a) ( x+2a)

= {x-a }^2 ( x+2a )

so it is divisible by ( x-a)^2

yes...it is a simple algebra basic question...adding and subtracting anything will add nothing to our equation..so add and subtract some terms as shown...so our equation becomes x^3+3a^3-a^3-6xa^2+3xa^2+3ax^2-3ax^2=(x^3-a^3-3x^2a+3xa^2)+(3a^3-6xa^2+3ax^2)=(x-a)^3+(a^2-2ax+x^2)3a=(x-a)^2{(x-a)+3a}=(x-a)^2(x+2a)...when it is divided by (x-a)^2 you will be left over with x+2a...so it is divisible...

Yes

this= x+2a