is 64x^15y^18 a perfect cube?
Would it be easier if we write \(64x^{15}y^{18}=2^6x^{15}y^{18} \) noting that all powers are divisible by three.
Yes, so it would be a difference of cubes because they are all divisible by 3?
A perfect cube will have all powers divisible by three, because, for example: \((5xy^2)^3=5^3x^3y^6\) the cube of an expression is obtained by multiplying the exponents by three.
Then how do i know if an expression is a difference of cubes?
Question is asking perfect cube where the exponents are all divisible by 3. Difference of cubes is a different concept, something like: \(\huge x^3-y^3 = (x-y)(x^2+xy+y^2)\) and is related to factorization.
ahhhh i think i get it, i was mistaking them to be together. Thank you
You're welcome! :)
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