Answer :
[tex]x^4-81=0 \\
(x^2)^2-9^2=0\\
(x^2-9)(x^2+9)=0\\
(x-3)(x+3)(x^2+9)=0\\
x=3 \vee x=-3[/tex]
so this is a difference of 2 perfect squares
so if
x^2-y^2 then it is equal to (x-y)(x+y)
so
x^4-81=[(x^2)^2]-[(9)^2]=(x^2-9)(x^2+9)
so we know if xy=0 then x or/and y=0 so assume that they are equal to zero
x^2+9=0
we can't do anything about this because
x^2=-9
and that is not possible in the real realm
x^2-9=0
yet another difference of 2 perfect squares
x^2-3^2=(x-3)(x+3)=0
set them to zero
x-3=0
x=3
x+3=0
x=-3
so x=3 or -3
so if
x^2-y^2 then it is equal to (x-y)(x+y)
so
x^4-81=[(x^2)^2]-[(9)^2]=(x^2-9)(x^2+9)
so we know if xy=0 then x or/and y=0 so assume that they are equal to zero
x^2+9=0
we can't do anything about this because
x^2=-9
and that is not possible in the real realm
x^2-9=0
yet another difference of 2 perfect squares
x^2-3^2=(x-3)(x+3)=0
set them to zero
x-3=0
x=3
x+3=0
x=-3
so x=3 or -3
