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Factorise: (2x+3y)³ - (2x-3y)³
Answers
Answered by
234
Use the identity: a³ - b³ = (a - b)(a² + ab + b²)
(2x+3y)³ - (2x-3y)³
= [(2x + 3y) - (2x-3y) ][(2x-3y)² + (2x - 3y)(2x + 3y) + (2x + 3y)² ]
= [ 2x + 3y - 2x + 3y ][(4x² + 9y² - 12xy) + (4x² - 9y²) + (4x² + 9y² + 12xy) ]
= [6y][ 4x² + 9y² - 12xy + 4x² - 9y² + 4x² + 9y² + 12xy ]
= [6y][ 4x² + 4x² + 4x² + 9y² + 9y² - 9y² - 12xy + 12xy ]
= [6y][ 12x² + 9y²]
= 6y×3×(4x² + 3y²)
= 18y(4x² + 3y²)
(2x+3y)³ - (2x-3y)³
= [(2x + 3y) - (2x-3y) ][(2x-3y)² + (2x - 3y)(2x + 3y) + (2x + 3y)² ]
= [ 2x + 3y - 2x + 3y ][(4x² + 9y² - 12xy) + (4x² - 9y²) + (4x² + 9y² + 12xy) ]
= [6y][ 4x² + 9y² - 12xy + 4x² - 9y² + 4x² + 9y² + 12xy ]
= [6y][ 4x² + 4x² + 4x² + 9y² + 9y² - 9y² - 12xy + 12xy ]
= [6y][ 12x² + 9y²]
= 6y×3×(4x² + 3y²)
= 18y(4x² + 3y²)
Anonymous:
You're not even watching the question.
what do you mean?
No..................................................................
It's the one that I wrote at first.
I replaced 54y square with a cube.
i know that but where did you get the 108 xy^2?
Shall I show the steps?
sure.
I'm typing it in the comments of the above answer.
check it there.
Answered by
54
I THINK THAT ANSWER IS
54y^3+72x^2y
process= [(2x)^3+3(2x)^29(3y)+3(2x)(3y)^2+(3y)^3]-[(2x)^3-3(2x)^2(3y)+3(2x)(3y)^2-(3y)^2]
= [8x^3+36x^2y+54xy^2+27y^3-8x^3+36x^2y-54xy^2+27y^3
= 54y^3+72x^2y
54y^3+72x^2y
process= [(2x)^3+3(2x)^29(3y)+3(2x)(3y)^2+(3y)^3]-[(2x)^3-3(2x)^2(3y)+3(2x)(3y)^2-(3y)^2]
= [8x^3+36x^2y+54xy^2+27y^3-8x^3+36x^2y-54xy^2+27y^3
= 54y^3+72x^2y
pls wait
sorry for wasting your time
yaaaaaaaa................
ggot itt!
thank you very much
whose should I mark as brainliest?
Nice!! you finally got it, so time is not wasted! ;)
whose shall I mark as brainliest?
can u help me in another problem?
The one which you think best should be marked as best!:)
I will try. give the link!
I will try. give the link!
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