For this problem, you can use Pascal's Triangle as a shortcut (http://mathematics.laerd.com/maths/binomial-theorem-intro.php), but you can also expand the long way:
(x+7y)3
= (x+7y) (x+7y) (x+7y)
= (x2 + 7xy + 7xy + 49y2) (x+7y)
= (x2 + 14xy + 49y2) (x + 7y)
= (x3 + 14x2y + 49xy2 + 7x2y + 98xy2 + 343y3)
= (x3 + 21x2y + 147xy2 + 343y3)
For this problem, you can use Pascal's Triangle as a shortcut (http://mathematics.laerd.com/maths/binomial-theorem-intro.php), but you can also expand the long way:
(x+7y)3
= (x+7y) (x+7y) (x+7y)
= (x2 + 7xy + 7xy + 49y2) (x+7y)
= (x2 + 14xy + 49y2) (x + 7y)
= (x3 + 14x2y + 49xy2 + 7x2y + 98xy2 + 343y3)
= (x3 + 21x2y + 147xy2 + 343y3)
The binomial theorem (for a cubic expansion) says:
(a + b)3 = a3 + 3a2b +3ab2 + b3
If you substitute x for a, and 7y for b in this you should get the result shown by kitty<3.
Hi kitty,
I'm sure your answer is correct - you rarely get any wrong - but that is not a binomial expansion.
(x+7y)3=(7y)3+3∗(7y)2(x)+3∗(7y)(x2)+(x)3Andyoucantakeitfromthere.
Actually, Melody, I think kitty<3 got exactly the same answer you did. She just took the coefficients to their "powers."
Yes, I know Kitty got it right. She almost never gets anything wrong!
But the question specifically asked the answerer to use the binomial theorem
If anyone want to have a look at the binomial theorum this clip seems pretty reasonable.
Oh, yes....I see now....she actually "cheated," didn't she???.......
Shame on you, kitty<3 !!!!!!!! (LOL!!!)