To find the value of [tex]\( x^3 + y^3 + 12xy \)[/tex] given that [tex]\( x + y = 4 \)[/tex], let's break it down step by step using algebraic identities.
First, recall the identity for the sum of cubes:
[tex]\[ x^3 + y^3 = (x+y)(x^2 - xy + y^2) \][/tex]
Given [tex]\( x + y = 4 \)[/tex], we can substitute:
[tex]\[ x^3 + y^3 = 4 \cdot (x^2 - xy + y^2) \][/tex]
Now, to further simplify [tex]\( x^2 - xy + y^2 \)[/tex], we use the square of the sum identity:
[tex]\[ (x + y)^2 = x^2 + 2xy + y^2 \][/tex]
Given [tex]\( x + y = 4 \)[/tex], squaring both sides gives:
[tex]\[ 16 = x^2 + 2xy + y^2 \][/tex]
We can solve for [tex]\( x^2 + y^2 \)[/tex] by rearranging the equation:
[tex]\[ x^2 + y^2 = 16 - 2xy \][/tex]
Next, we substitute [tex]\( x^2 + y^2 \)[/tex] back into our previous expression:
[tex]\[ x^3 + y^3 = 4 \cdot (x^2 - xy + y^2) = 4 \cdot [(16 - 2xy) - xy] \][/tex]
[tex]\[ x^3 + y^3 = 4 \cdot (16 - 3xy) \][/tex]
[tex]\[ x^3 + y^3 = 64 - 12xy \][/tex]
Now, to find [tex]\( x^3 + y^3 + 12xy \)[/tex]:
[tex]\[ x^3 + y^3 + 12xy = (64 - 12xy) + 12xy \][/tex]
[tex]\[ x^3 + y^3 + 12xy = 64 \][/tex]
Therefore, the value of [tex]\( x^3 + y^3 + 12xy \)[/tex] is [tex]\( \boxed{64} \)[/tex].
