Need A Little Help (Easy (I think))

A

approcotE

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Apr 28, 2020
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Hey, this is my first post here!

I've been having an issue with this question

I'm pretty sure I could do a sneaky move and complete it quickly but I'm curious if anyone knows the 'correct' answer

:)

Math Question #14.png
 
Show us your "sneaky move"; it may be exactly what I'd do. This is not a straightforward application of the binomial square formula, but involves moving forward and backward.

I myself would first fill in the last blank, then the first, then the second. But you could fill the first blank quickly, too.

There is one correct answer, but there could be multiple valid methods for finding it. Don't look for "the" perfect way; just do something and see what happens.
 
I finished the question I guess without using my 'sneaky move' lol.

Idk how it happened but it was really satisfying!
 
I would replace that "blank" on the left by a letter, say "a", and do the square:
\(\displaystyle (a+ 8y^3)^2= a^2+ 16ay^3+ 64y^3\)= ____+ \(\displaystyle 48x^4y^3\)+ _____.

So you want \(\displaystyle 16ay^3= 48x^4y^3\). What must "a" be?
 
That's your answer, not what you did; so I can't tell whether there's any more to discuss.

But it is correct, so we can leave it at that.
 
HallsofIvy said:
I would replace that "blank" on the left by a letter, say "a", and do the square:
\(\displaystyle (a+ 8y^3)^2= a^2+ 16ay^3+ 64y^3\)= ____+ \(\displaystyle 48x^4y^3\)+ _____.

So you want \(\displaystyle 16ay^3= 48x^4y^3\). What must "a" be?
Click to expand...
Since the answer has been given, but without an explanation:
For \(\displaystyle 16ay^3= 48x^4y^3\), a must be \(\displaystyle a= 48x^4y^3/16y^3= 3x^4\).

Then \(\displaystyle (a+ 8y^3)^2= (3x^4+ 8y^3)= (3x^4)^2+ 2(3x^4)(8y^3)+ (8y^3)^2\)
\(\displaystyle = 9x^8+ 48x^4y^3+ 64y^6\).
 
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