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How do you factor 8x^3*y^6 + 27? | HIX Tutor

Sum of cubes: <mathjax>#(a^3+b^3)=(a+b)(a^2-ab+b^2)#</mathjax>
Since <mathjax>#8x^3y^6+27=(2xy^2)^3+(3)^3#</mathjax>, 
<mathjax>#a=2xy^2#</mathjax> 
<mathjax>#b=3#</mathjax>
<mathjax>#8x^3y^6+27=(2xy^2+3)(4x^2y^4-6xy^2+9)#</mathjax>

Sum of cubes: <mathjax>#(a^3+b^3)=(a+b)(a^2-ab+b^2)#
Since <mathjax>#8x^3y^6+27=(2xy^2)^3+(3)^3#</mathjax>, 
<mathjax>#a=2xy^2#</mathjax> 
<mathjax>#b=3#
<mathjax>#8x^3y^6+27=(2xy^2+3)(4x^2y^4-6xy^2+9)#

Ethan Adkins

To factor the expression \(8x^3y^6 + 27\), you can use the sum of cubes formula, which states that \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). In this case, \(8x^3y^6\) can be expressed as \((2xy^2)^3\) and \(27\) can be expressed as \(3^3\). So, the expression can be factored as \((2xy^2 + 3)(4x^2y^4 - 6xy^2 + 9)\).

How do you factor #8x^3*y^6 + 27#?