Question 196635
{{{a^3x^3 + 64 = 0}}} Start with the given equation.



{{{(ax)^3 + 64 = 0}}} Rewrite {{{a^3x^3}}} as {{{(ax)^3}}}



{{{(ax)^3 + 4^3 = 0}}} Rewrite {{{64}}} as {{{(4)^3}}}



{{{(ax+4)((ax)^2-(ax)(4)+(4)^2) = 0}}} Factor the left side using the sum of cubes formula



Recall, the sum of cubes formula is {{{A^3+B^3=(A+B)(A^2-AB+B^2)}}}



{{{(ax+4)(a^2x^2-4ax+16) = 0}}} Multiply and simplify



{{{ax+4=0}}} or {{{a^2x^2-4ax+16=0}}} Use the zero product property to break up the factors



-----------------------------------------------------


Let's solve the first equation: {{{ax+4=0}}}



{{{ax+4=0}}} Start with the first equation.



{{{ax=-4}}} Subtract 4 from both sides.



{{{x=-4/a}}} Divide both sides by "a" to isolate "x".



So the first solution is {{{x=-4/a}}}


-----------------------------------------------------


Now let's solve the second equation: {{{a^2x^2-4ax+16=0}}}



{{{a^2x^2-4ax+16=0}}} Start with the second equation.



Notice we have a quadratic in the form of {{{Ax^2+Bx+C}}} where {{{A=a^2}}}, {{{B=-4a}}}, and {{{C=16}}}



Let's use the quadratic formula to solve for "x"



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(-4a) +- sqrt( (-4a)^2-4(a^2)(16) ))/(2(a^2))}}} Plug in {{{A=a^2}}}, {{{B=-4a}}}, and {{{C=16}}}



{{{x = (4a +- sqrt( (-4a)^2-4(a^2)(16) ))/(2(a^2))}}} Negate -4a to get 4a



{{{x = (4a +- sqrt( 16a^2-4(a^2)(16) ))/(2(a^2))}}} Square -4a to get {{{16a^2}}}



{{{x = (4a +- sqrt( 16a^2-64a^2 ))/(2a^2)}}} Multiply



{{{x = (4a +- sqrt( -48a^2 ))/(2a^2)}}} Combine like terms.



{{{x = (4a +- sqrt( -1*16*3*a^2 ))/(2a^2)}}} Factor -48 into {{{-1*16*3}}}



{{{x = (4a +- sqrt(-1)*sqrt(16)*sqrt(3)*sqrt(a^2) )/(2a^2)}}} Break up the square root.



{{{x = (4a +- i*4*sqrt(3)*a )/(2a^2)}}} Simplify the square roots and replace {{{sqrt(-1)}}} with "i".



Note: {{{i=sqrt(-1)}}}



{{{x = (4a +- 4a*sqrt(3)*i )/(2a^2)}}} Rearrange the terms.



{{{x = (4a + 4a*sqrt(3)*i )/(2a^2)}}} or {{{a = (4a - 4a*sqrt(3)*i )/(2a^2)}}} Break up the "plus/minus"



{{{x = (2 + 2*sqrt(3)*i )/a}}} or {{{x = (2 - 2*sqrt(3)*i )/a}}} Reduce



So the next two solutions are {{{x = (2 + 2*sqrt(3)*i )/a}}} or {{{x = (2 - 2*sqrt(3)*i )/a}}}



================================================


Answer:


So the three solutions are {{{x=-4/a}}}, {{{x = (2 + 2*sqrt(3)*i )/a}}} or {{{x = (2 - 2*sqrt(3)*i )/a}}}