# How do you FOIL  (x^3 – x^2)(x^3 + x^2)?

Mar 3, 2018

$\left({x}^{3} - {x}^{2}\right) \left({x}^{3} + {x}^{2}\right) = {x}^{6} - {x}^{4}$

#### Explanation:

FOIL is a mnemonic to help remember all of the combinations you need to multiply and add when finding the product of two binomials.

In our example, we find:

$\left({x}^{3} - {x}^{2}\right) \left({x}^{3} + {x}^{2}\right) = {\overbrace{\left({x}^{3}\right) \left({x}^{3}\right)}}^{\text{First"+overbrace((x^3)(x^2))^"Outside"+overbrace((-x^2)(x^3))^"Inside"+overbrace((-x^2)(x^2))^"Last}}$

$\textcolor{w h i t e}{\left({x}^{3} - {x}^{2}\right) \left({x}^{3} + {x}^{2}\right)} = {x}^{6} + \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{5}}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{5}}}} - {x}^{4}$

$\textcolor{w h i t e}{\left({x}^{3} - {x}^{2}\right) \left({x}^{3} + {x}^{2}\right)} = {x}^{6} - {x}^{4}$

In fact, in this particular example, we might note that the multiplication takes the form $\left(A - B\right) \left(A + B\right)$ with $A = {x}^{3}$ and $B = {x}^{2}$.

In general, we have:

$\left(A - B\right) \left(A + B\right) = {A}^{2} - {B}^{2}$

This is known as the difference of squares identity.

In our particular example:

$\left({x}^{3} - {x}^{2}\right) \left({x}^{3} + {x}^{2}\right) = {\left({x}^{3}\right)}^{2} - {\left({x}^{2}\right)}^{2} = {x}^{6} - {x}^{4}$

Mar 3, 2018

${x}^{6} - {x}^{4}$, or alternatively ${x}^{2} \left({x}^{3} - {x}^{2}\right)$ (factored form)

#### Explanation:

FOIL is extremely useful in multiplying binomials. To be able to use it, let's understand what it means first:

FOIL stands for Firsts, Outsides, Insides, Lasts, meaning we multiply the first terms, outside terms, inside terms, and last terms, respectively.

In $\left({x}^{3} - {x}^{2}\right) \left({x}^{3} + {x}^{2}\right)$, our terms are as follows:

• Firsts $\left({x}^{3} \cdot {x}^{3}\right) = {x}^{6}$
• Outsides $\left({x}^{3} \cdot {x}^{2}\right) = {x}^{5}$
• Insides $\left(- {x}^{2} \cdot {x}^{3}\right) = - {x}^{5}$
• Lasts $\left(- {x}^{2} \cdot {x}^{2}\right) = - {x}^{4}$

Thus, we have:

${x}^{6} + {x}^{5} - {x}^{5} - {x}^{4}$

The middle terms will cancel out (as expected, because the original problem is in the difference of squares format), and we get:

${x}^{6} - {x}^{4}$

We can factor out an ${x}^{2}$ (which is essentially dividing by ${x}^{2}$). We get:

${x}^{2} \left({x}^{3} - {x}^{2}\right)$

Mar 3, 2018

the answer is ${x}^{6}$-${x}^{4}$
(${x}^{3}$-${x}^{2}$)(${x}^{3}$+${x}^{2}$)=
=${x}^{3}$${x}^{3}$ + ${x}^{3}$${x}^{2}$ - ${x}^{2}$${x}^{3}$ - ${x}^{2}$${x}^{2}$=
=${x}^{6}$-${x}^{4}$