An interesting fact:
a^3+b^3=(a+b)(a^2-ab+b^2)a3+b3=(a+b)(a2−ab+b2)
In x^3+8x3+8, a^3=x^3a3=x3 and b^3=8b3=8
Let's solve for aa and bb.
=>a^3=x^3⇒a3=x3
=>root [3] (a^3)= root[3] (x^3)⇒3√a3=3√x3
=>a= x⇒a=x
Now for bb.
=>b^3=8⇒b3=8
=>root [3] (b^3)= root[3] (8)⇒3√b3=3√8
=>b= 2⇒b=2
Plug these values into our equation.
x^3+2^3=(x+2)(x^2-2x+2^2)x3+23=(x+2)(x2−2x+22)
(x+2)(x^2-2x+4)(x+2)(x2−2x+4) This is our answer!
If you want to factor this further, we let x^2-2x+4=0x2−2x+4=0 and solve the equation.
x^2-2x+4=0x2−2x+4=0 Use the quadratic formula:
(-b+-sqrt(b^2-4(a)(c)))/(2(a))−b±√b2−4(a)(c)2(a)
Here, a=1a=1, b=-2b=−2, and c=4c=4
x=(-(-2)+-sqrt((-2)^2-4(1)(4)))/(2(1))x=−(−2)±√(−2)2−4(1)(4)2(1)
x=(2+-sqrt(4-16))/(2)x=2±√4−162
x=(2+-sqrt(-12))/(2)x=2±√−122
x=(2+-2isqrt(3))/(2)x=2±2i√32
x=1+-isqrt(3)x=1±i√3
(x+2)(x-(1+isqrt3))(x-(1-isqrt3))(x+2)(x−(1+i√3))(x−(1−i√3))
(x+2)(x-1-isqrt3))(x-1+isqrt3))(x+2)(x−1−i√3))(x−1+i√3))
Therefore, our factored form, in this case, would be
(x+2)(x-1-isqrt3))(x-1+isqrt3))(x+2)(x−1−i√3))(x−1+i√3))
