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

2 Answers
Apr 5, 2018

Answer:

#345#

Explanation:

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

#x^3+2y^3#

#7^3+2·1^3=345#

Apr 5, 2018

Answer:

#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#.


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