# How do you solve x^3 + 2y^3 for x = 7 & y = 1?

Apr 5, 2018

$345$

#### Explanation:

Just by replaicing the values in $x$ and $y$:

${x}^{3} + 2 {y}^{3}$

7^3+2·1^3=345

Apr 5, 2018

$345$

Here's how I did it:

#### Explanation:

Your expression is ${x}^{3} + 2 {y}^{3}$ and you know that $x = 7$ and $y = 1$. We just plug in these values for $x$ and $y$, and then we simplify:

${\left(7\right)}^{3} + 2 {\left(1\right)}^{3}$

${7}^{3}$ is the same thing as $\left(7\right) \left(7\right) \left(7\right)$, or $343$.

Now, for $2 {\left(1\right)}^{3}$, we have to look this order of operations called PEMDAS. This stands for:
P arentheses
E xponents
MD Multiplication/Division
In your expression $2 {\left(1\right)}^{3}$, the first thing we need to do is take care of the exponent. (1)^3 is the same thing as $\left(1\right) \left(1\right) \left(1\right)$, which is just $1$.
So now we do $2 \left(1\right)$, which is $2$.
Finally we add $2$ with $343$ (from earlier) to get the final simplification of $345$.