How to factor the sum of two cubes
The formula we use to factor a binomial which is the sum of two perfect cubes
In this lesson we’ll look at how to recognize a sum of two cubes and then use a formula to factor it.
We’ll know when we have a sum of cubes because we’ll have two perfect cubes separated by addition. When that’s the case, we can take the cube (third) root of each term and use a formula to factor.
The formula for the sum of two cubes is
???a^3+b^3=(a+b)(a^2-ab+b^2)???
How to factor the sum of two cubes
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Factoring the sum of two perfect squares, step-by-step
Example
Factor the polynomial.
???125x^3+512y^3z^9???
First check to see if each term is a perfect cube.
???\sqrt[3]{125x^3}=5x???
???\sqrt[3]{512y^3z^9}=8yz^3???
Both terms are perfect cubes, so we can use the cube plus a cube formula to factor. The formula is
???a^3+b^3=(a+b)(a^2-ab+b^2)???
In this case ???a=5x??? and ???b=8yz^3???, so we can plug these into our formula and get
???(5x+8yz^3)\left((5x)^2-(5x)(8yz^3)+(8yz^3)^2\right)???
???(5x+8yz^3)(25x^2-40xyz^3+64y^2z^6)???
We can check our work by distributing each term in the binomial factor over each term in the trinomial factor.
???5x(25x^2-40xyz^3+64y^2z^6)+8yz^3(25x^2-40xyz^3+64y^2z^6)???
???5x(25x^2)-5x(40xyz^3)+5x(64y^2z^6)+8yz^3(25x^2)-8yz^3(40xyz^3)+8yz^3(64y^2z^6)???
???125x^3-200x^2yz^3+320xy^2z^6+200x^2yz^3-320xy^2z^6+512y^3z^9???
???125x^3-200x^2yz^3+200x^2yz^3+320xy^2z^6-320xy^2z^6+512y^3z^9???
???125x^3+512y^3z^9???
Let’s do another example.
Example
Factor the expression.
???729h^{30}j^9+27m^{15}n^3???
First check to see if each term is a perfect cube.
???\sqrt[3]{729h^{30}j^9}=9h^{10}j^3???
???\sqrt[3]{27m^15n^3}=3m^5n???
Both terms are perfect cubes, so we can use the formula for factoring the sum of perfect cubes.
???a^3+b^3=(a+b)(a^2-ab+b^2)???
In this case,
???a=9h^{10}j^3???
???b=3m^5n???
We use the sum of cubes formula to get
???(9h^{10}j^3+3m^5n)((9h^{10}j^3)^2-(9h^{10}j^3)(3m^5n)+(3m^5n)^2)???
???(9h^{10}j^3+3m^5n)(81h^{20}j^6-27h^{10}j^3m^5n+9m^{10}n^2)???