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#1**+29 **

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)

= (x^{2} + 7xy + 7xy + 49y^{2}) (x+7y)

= (x^{2 }+ 14xy + 49y^{2}) (x + 7y)

= (x^{3} + 14x^{2}y + 49xy^{2} + 7x^{2}y + 98xy^{2} + 343y^{3})

= (x^{3} + 21x^{2}y + 147xy^{2} + 343y^{3})

kitty<3 Jun 6, 2014

#1**+29 **

Best Answer

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)

= (x^{2} + 7xy + 7xy + 49y^{2}) (x+7y)

= (x^{2 }+ 14xy + 49y^{2}) (x + 7y)

= (x^{3} + 14x^{2}y + 49xy^{2} + 7x^{2}y + 98xy^{2} + 343y^{3})

= (x^{3} + 21x^{2}y + 147xy^{2} + 343y^{3})

kitty<3 Jun 6, 2014

#2**+8 **

The binomial theorem (for a cubic expansion) says:

(a + b)^{3} = a^{3} + 3a^{2}b +3ab^{2} + b^{3}

If you substitute x for a, and 7y for b in this you should get the result shown by kitty<3.

Alan Jun 6, 2014

#3**+8 **

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)(x^2)+(x)^3

And you can take it from there.$$

Melody Jun 6, 2014

#4**+3 **

Actually, Melody, I think kitty<3 got exactly the same answer you did. She just took the coefficients to their "powers."

CPhill Jun 6, 2014

#5**+3 **

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.

Melody Jun 6, 2014

#6**+3 **

Oh, yes....I see now....she actually "cheated," didn't she???.......

Shame on you, kitty<3 !!!!!!!! (LOL!!!)

CPhill Jun 6, 2014