Factoring the sum and difference of two cubes uses special patterns for expressions like \(a^3+b^3\) and \(a^3-b^3\). The key is to identify each cube, write it as \(a^3\) and \(b^3\), and then apply the correct formula. These problems include basic cube factoring, expressions with variables, problems with a greatest common factor, and challenge problems that require factoring completely.
Notes
Practice Problems
Factor
\(\textbf{1)}\) \( x^3-1000 \)
The answer is \( (x-10)(x^2+10x+100) \)
\(\,\,\,\,\,\,x^3-1000\)
\(\,\,\,\,\,\,x^3-10^3\)
\(\,\,\,\,\,\,a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(\,\,\,\,\,\,a=x,\quad b=10\)
\(\,\,\,\,\,\,\left(x-10\right)\left(x^2+10x+100\right)\)
\(\textbf{2)}\) \( x^3-1 \)
The answer is \( (x-1)(x^2+x+1) \)
\(\,\,\,\,\,\,x^3-1\)
\(\,\,\,\,\,\,x^3-1^3\)
\(\,\,\,\,\,\,a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(\,\,\,\,\,\,a=x,\quad b=1\)
\(\,\,\,\,\,\,\left(x-1\right)\left(x^2+x+1\right)\)
\(\textbf{3)}\) \( y^6+1 \)
The answer is \( (y^2+1)(y^4-y^2+1) \)
\(\,\,\,\,\,\,y^6+1\)
\(\,\,\,\,\,\,\left(y^2\right)^3+1^3\)
\(\,\,\,\,\,\,a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\,\,\,\,\,\,a=y^2,\quad b=1\)
\(\,\,\,\,\,\,\left(y^2+1\right)\left(y^4-y^2+1\right)\)
\(\textbf{4)}\) \( 16x^4-54x \)
The answer is \( 2x(2x-3)(4x^2+6x+9) \)
\(\text{Step 1: Factor out a GCF}\)
\(\,\,\,\,\,\,16x^4-54x=2x\left(8x^3-27\right)\)
\(\text{Step 2: Use the difference of two cubes equation}\)
\(\,\,\,\,\,\,a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(\,\,\,\,\,\,a=\sqrt[3]{8x^3}=2x,\quad b=\sqrt[3]{27}=3\)
\(\,\,\,\,\,\,8x^3-27=\left(2x-3\right)\left(4x^2+6x+9\right)\)
\(\text{Step 3: Bring down the GCF}\)
\(\,\,\,\,\,\,2x\left(2x-3\right)(4x^2+6x+9)\)
\(\textbf{5)}\) \( w^3-27y^9 \)
The answer is \( (w-3y^3)(w^2+3wy^3+9y^6) \)
\(\,\,\,\,\,\,w^3-27y^9\)
\(\,\,\,\,\,\,w^3-\left(3y^3\right)^3\)
\(\,\,\,\,\,\,a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(\,\,\,\,\,\,a=w,\quad b=3y^3\)
\(\,\,\,\,\,\,\left(w-3y^3\right)\left(w^2+3wy^3+9y^6\right)\)
\(\textbf{6)}\) \( x^3+64 \)
The answer is \( (x+4)(x^2-4x+16) \)
\(\text{Use the sum of two cubes equation}\)
\(\,\,\,\,\,\,a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\,\,\,\,\,\,a=\sqrt[3]{x^3}=x,\quad b=\sqrt[3]{64}=4\)
\(\,\,\,\,\,\,x^3+64=x^3+4^3\)
\(\,\,\,\,\,\,\left(x+4\right)\left(x^2-4x+16\right)\)
\(\textbf{7)}\) \( w^{12}+64y^3 \)
The answer is \( (w^4+4y)(w^8-4w^4 y+16y^2) \)
\(\,\,\,\,\,\,w^{12}+64y^3\)
\(\,\,\,\,\,\,\left(w^4\right)^3+\left(4y\right)^3\)
\(\,\,\,\,\,\,a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\,\,\,\,\,\,a=w^4,\quad b=4y\)
\(\,\,\,\,\,\,\left(w^4+4y\right)\left(w^8-4w^4y+16y^2\right)\)
\(\textbf{8)}\) \( x^3-8 \)
The answer is \( (x-2)(x^2+2x+4) \)
\(\,\,\,\,\,\,x^3-8\)
\(\,\,\,\,\,\,x^3-2^3\)
\(\,\,\,\,\,\,a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(\,\,\,\,\,\,a=x,\quad b=2\)
\(\,\,\,\,\,\,\left(x-2\right)\left(x^2+2x+4\right)\)
\(\textbf{9)}\) \(27x^3+8\)
The answer is \(\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(\,\,\,\,\,\,27x^3+8\)
\(\,\,\,\,\,\,\left(3x\right)^3+2^3\)
\(\,\,\,\,\,\,a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\,\,\,\,\,\,a=3x,\quad b=2\)
\(\,\,\,\,\,\,\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(\textbf{10)}\) \(64a^3-b^3\)
The answer is \(\left(4a-b\right)\left(16a^2+4ab+b^2\right)\)
\(\,\,\,\,\,\,64a^3-b^3\)
\(\,\,\,\,\,\,\left(4a\right)^3-b^3\)
\(\,\,\,\,\,\,a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(\,\,\,\,\,\,a=4a,\quad b=b\)
\(\,\,\,\,\,\,\left(4a-b\right)\left(16a^2+4ab+b^2\right)\)
\(\textbf{11)}\) \(125m^3+27n^3\)
The answer is \(\left(5m+3n\right)\left(25m^2-15mn+9n^2\right)\)
\(\,\,\,\,\,\,125m^3+27n^3\)
\(\,\,\,\,\,\,\left(5m\right)^3+\left(3n\right)^3\)
\(\,\,\,\,\,\,a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\,\,\,\,\,\,a=5m,\quad b=3n\)
\(\,\,\,\,\,\,\left(5m+3n\right)\left(25m^2-15mn+9n^2\right)\)
\(\textbf{12)}\) \(8x^6-1\)
The answer is \(\left(2x^2-1\right)\left(4x^4+2x^2+1\right)\)
\(\,\,\,\,\,\,8x^6-1\)
\(\,\,\,\,\,\,\left(2x^2\right)^3-1^3\)
\(\,\,\,\,\,\,a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(\,\,\,\,\,\,a=2x^2,\quad b=1\)
\(\,\,\,\,\,\,\left(2x^2-1\right)\left(4x^4+2x^2+1\right)\)
\(\textbf{13)}\) \(27p^6+64\)
The answer is \(\left(3p^2+4\right)\left(9p^4-12p^2+16\right)\)
\(\,\,\,\,\,\,27p^6+64\)
\(\,\,\,\,\,\,\left(3p^2\right)^3+4^3\)
\(\,\,\,\,\,\,a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\,\,\,\,\,\,a=3p^2,\quad b=4\)
\(\,\,\,\,\,\,\left(3p^2+4\right)\left(9p^4-12p^2+16\right)\)
\(\textbf{14)}\) \(250x^4-16x\)
The answer is \(2x\left(5x-2\right)\left(25x^2+10x+4\right)\)
\(\,\,\,\,\,\,250x^4-16x\)
\(\,\,\,\,\,\,2x\left(125x^3-8\right)\)
\(\,\,\,\,\,\,2x\left(\left(5x\right)^3-2^3\right)\)
\(\,\,\,\,\,\,a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(\,\,\,\,\,\,2x\left(5x-2\right)\left(25x^2+10x+4\right)\)
\(\textbf{15)}\) \(64u^9+v^3\)
The answer is \(\left(4u^3+v\right)\left(16u^6-4u^3v+v^2\right)\)
\(\,\,\,\,\,\,64u^9+v^3\)
\(\,\,\,\,\,\,\left(4u^3\right)^3+v^3\)
\(\,\,\,\,\,\,a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\,\,\,\,\,\,a=4u^3,\quad b=v\)
\(\,\,\,\,\,\,\left(4u^3+v\right)\left(16u^6-4u^3v+v^2\right)\)
Challenge Problems
\(\textbf{16)}\) Factor \( x^6-64 \) completely
The answer is \( \left(x+2\right)\left(x-2\right)\left(x^2+2x+4\right)\left(x^2-2x+4\right) \)
\(\,\,\,\,\,\,x^6-64\)
\(\,\,\,\,\,\,\left(x^3\right)^2-8^2\)
\(\,\,\,\,\,\,\left(x^3-8\right)\left(x^3+8\right)\)
\(\,\,\,\,\,\,\left(x^3-2^3\right)\left(x^3+2^3\right)\)
\(\,\,\,\,\,\,\left(x-2\right)\left(x^2+2x+4\right)\left(x+2\right)\left(x^2-2x+4\right)\)
\(\textbf{17)}\) Factor \( x^6-1 \) completely
The answer is \( \left(x+1\right)\left(x-1\right)\left(x^2+x+1\right)\left(x^2-x+1\right) \)
\(\,\,\,\,\,\,x^6-1\)
\(\,\,\,\,\,\,\left(x^3\right)^2-1^2\)
\(\,\,\,\,\,\,\left(x^3-1\right)\left(x^3+1\right)\)
\(\,\,\,\,\,\,\left(x^3-1^3\right)\left(x^3+1^3\right)\)
\(\,\,\,\,\,\,\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(\textbf{18)}\) Factor \(8x^9-27y^6\) completely
The answer is \(\left(2x^3-3y^2\right)\left(4x^6+6x^3y^2+9y^4\right)\)
\(\,\,\,\,\,\,8x^9-27y^6\)
\(\,\,\,\,\,\,\left(2x^3\right)^3-\left(3y^2\right)^3\)
\(\,\,\,\,\,\,a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(\,\,\,\,\,\,a=2x^3,\quad b=3y^2\)
\(\,\,\,\,\,\,\left(2x^3-3y^2\right)\left(4x^6+6x^3y^2+9y^4\right)\)
\(\textbf{19)}\) Factor \(64x^6-1\) completely
The answer is \(\left(2x-1\right)\left(2x+1\right)\left(16x^4+4x^2+1\right)\)
\(\,\,\,\,\,\,64x^6-1\)
\(\,\,\,\,\,\,\left(4x^2\right)^3-1^3\)
\(\,\,\,\,\,\,\left(4x^2-1\right)\left(16x^4+4x^2+1\right)\)
\(\,\,\,\,\,\,4x^2-1=\left(2x-1\right)\left(2x+1\right)\)
\(\,\,\,\,\,\,\left(2x-1\right)\left(2x+1\right)\left(16x^4+4x^2+1\right)\)
\(\textbf{20)}\) Factor \(x^{12}-y^{12}\) completely
The answer is \(\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)
\(\,\,\,\,\,\,x^{12}-y^{12}\)
\(\,\,\,\,\,\,\left(x^6\right)^2-\left(y^6\right)^2\)
\(\,\,\,\,\,\,\left(x^6-y^6\right)\left(x^6+y^6\right)\)
\(\,\,\,\,\,\,x^6-y^6=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(\,\,\,\,\,\,x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(\,\,\,\,\,\,x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(\,\,\,\,\,\,x^6+y^6=\left(x^2\right)^3+\left(y^2\right)^3\)
\(\,\,\,\,\,\,x^6+y^6=\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)\)
\(\,\,\,\,\,\,\left(x^4-x^2y^2+y^4\right)=\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)
\(\,\,\,\,\,\,\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^2+xy+y^2\right)^2\left(x^2-xy+y^2\right)^2\)
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