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factor: 64- a^2+2ab-b^2
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64- a^2+2ab-b^2 = 64 - (a^2 - 2ab + b^2) Apply the identity \((a\pm b)^2 = a^2 \pm 2ab + b^2\) = 64 - ( a-b)^2 = 8^2 - (a-b)^2 Apply the identity \(a^2 - b^2 = (a+b)(a-b)\) = [8-(a-b)][8+(a-b)] Simplify the things in the bracket, can you do it?
thank you so much! so it would be (8-a+b) and (8+a-b)?
Yup :)
you're amazing! can I ask you one more?
I'm not, but you can ask :)
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factor completely: 243x^5-y^15
Hmm... not sure for this one :| \[243x^5-y^{15} \]\[=(3x)^5-(y^{3})^5 \]\[=(3x-y^3)[(3x)^4+(3x)^3(y^3)+(3x)^2(y^3)^2+(3x)(y^3)^3+(y^3)^4]\]\[=(3x-y^3)(81x^4+27x^3y^3+9x^2y^6+3xy^9+y^{12})\]
i did it the same way! perfect thank you thank you!
welcome :)
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