Question

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

Answer 1

`345`

Explanation:

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

`x^3+2y^3`

`7^3+2·1^3=345`

Answer 2

`345`

Here's how I did it:

Explanation:

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

`(7)^3 + 2(1)^3`

`7^3` is the same thing as `(7)(7)(7)`, or `343`.

Now, for `2(1)^3`, we have to look this order of operations called PEMDAS. This stands for:
P arentheses
E xponents
MD Multiplication/Division
AS Addition/Subtraction

In your expression `2(1)^3`, the first thing we need to do is take care of the exponent. (1)^3 is the same thing as `(1)(1)(1)`, which is just `1`.

So now we do `2(1)`, which is `2`.

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Finally we add `2` with `343` (from earlier) to get the final simplification of `345`.