factorize (2x-y)^{6}-(2x+y)^{6}
I think Larry's method is simple and clear enough
factorize (2x-y)^{6}-(2x+y)^{6}
I think Larry's method is simple and clear enough
The expression (2x-y)^{6}-(2x+y)^{6} is a difference of squares since it is [(2x-y)^{3}]^{2 }- [(2x-y)^{3}]^{2}. Now recall the difference of squares formula: a^{2} - b^{2} = (a+b)(a-b). Here, a = (2x-y)^{3} and b = (2x+y)^{3}, so we can apply the diffference of squares formula.
[(2x-y)^{3}]^{2} - [(2x+y)^{3}]^{2} = [(2x-y)^{3} +(2x+y)^{3}][(2x-y)^{3}-(2x+y)^{3}]
The right side involves cubing binomials. The general formula for cubing a binomial is (a+b)^{3} = a^{3} +3a^{2}b + 3ab^{2} + b^{3}. If you replace b with -b, the formula becomes
(a-b)^{3}=a^{3} - 3a^{2}b + 3ab^{2} - b^{3}. Here, a is 2x and b is y.
So (2x-y)^{3} = (2x)^{3} - 3(2x)^{2}y + 3(2x)y^{2} - y^{3 }= 8x^{3} - 12x^{2}y + 6xy^{2 }- y^{3}
and (2x+y)^{3} = (2x)^{3} + 3(2x)^{2}y + 3(2x)y^{2} + y^{3} = 8x^{3} + 12x^{2}y + 6xy^{2} + y^{3}.
The right side, again, is [(2x-y)^{3} +(2x+y)^{3}][(2x-y)^{3}-(2x+y)^{3}]
When you simplify the first factor (2x-y)^{3} +(2x+y)^{3}, line up the above expressions and two pairs of terms cancel, leaving 16x^{3} +12xy^{2}.
When you simplify the second factor (2x-y)^{3} - (2x+y)^{3}, line up the above expressions and two pairs of terms cancel, leaving -24x^{2}y - 2y^{3}.
Therefore, [(2x-y)^{3} +(2x+y)^{3}][(2x-y)^{3}-(2x+y)^{3}] = (16x^{3} +12xy^{2})(-24x^{2}y - 2y^{3})
The first factor has a common factor of 4x and the second factor has a common factor of -2y. Pull these out to simplify further.
(16x^{3} +12xy^{2})(-24x^{2}y - 2y^{3}) = 4x(4x^{2} + 3y^{2})(-2y)(12x^{2} + y^{2}) and these factors cannot be simplified further. Combine the first and third factor and put in front to clean it up.
= -8xy(4x^{2} + 3y^{2})(12x^{2} + y^{2})
Giridhar,
Expanding polynomials when the exponent is greater than 3 are complicated and it is best to use Pascal's triangle. The x exponent in the first term will be 6 and then descend from there as the exponent on the y will increase. The coefficients on each term will be the numbers in the (n+1) row of Pascal's Triangle (you can find this on the internet or your math book). Be careful though because there is a 2 infront of the x so terms with an x need to be multiplied by two raised to the exponent of the x term..
Also notice the signs on each of the expressions and that you are subtracting one polynomial from another. Half of the terms are going to cross out because you are subracting identical terms. Based on my calculations, only the terms with odd exponents remain.
Good luck!