Answer:
Explanation:
Let's start by pulling out some common factors. There are lots of as and bs floating around. Let's get them out where we can.
a^2 b^2 looks like it is common to all three terms.
a^2*b^2 * [ (9/25) a^4 - a^2*b^2 + (25/36) b^4]
If that is going to form a binomial that can be squared to get those three terms above, then the binomial will be related to the square root of the first and third term.
Square root of First term: sqrt((9/25) a^4) = 3/5 * a^2
Square root of Third term: sqrt((25/36) b^4) = 5/6 b^2
So if this is going to work at all the binomial will look like
(3/5 a^2 - 5/6 b^2 ) Plus the common factor we pulled out which we will deal with later.
And when it is squared it will look like
a^2 b^2 (3/5 a^2 - 5/6 b^2) ^2
Is this right? There's only one way to find out. We have to square this binomial and see if we get the same answer we started with which is the trinomial beginning with (9/25)a^6b^2 ...
So square the binomial
a^2b^2 [ 9/25 a^4 - 2(3/5a^2)(5/6b^2) + 25/36b^4 ] The first and third terms look pretty good. It's the middle term that's going to give the problem. Fortunately cancellation will come to our rescue.
2*3/5*a^2 *5/6 b^2
6/5 a^2 * 5/6 b^2 = a^2* b^2.
So what you have now is
a^2*b^2[9/25 a^4 - a^2*b^2 + 25/36 b^4] which is pretty much what we started with. I'll leave you with putting a^2b^2 into the brackets to get the absolute starting trinomial.