Question 213777
1. x^3+216=0


{{{x^3+216-216=0-216}}}   Subtracting 216 from both sides of equation


{{{x^3=-216}}}


{{{x=-6}}}  Answer


2. 5x^3+30x^2+45x=0


{{{x(5x^2+30x+45)=0}}}


{{{x(5x+ 15)(x+3) = 0 }}}


where we have x=0, 5x+15=0 and x+3=0.  Solving yields


x=0, x=-3 and x=-3, or solve quadratic equation using following steps,


*[invoke quadratic "x", 5, 30, 45 ]



3. 1/x^2+8/x+15=0  Multiply by x^2 to get rid of denominator


{{{x^2(1/x^2+8/x+15)*x^2=0}}}


{{{1+8x+15x^2=0}}}


{{{(5x+1)(3x+1)=0}}}


where we have 5x+1=0  and 3x+1=0.  Solving yields


So x=(-1/5) and x=(-1/3)
  

We can also use the quadratic formula as follows:  


*[invoke quadratic "x", 15, 8, 1 ]


I hope the above steps were helpful. 

 
And good luck in your studies!


For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.


Respectfully,
Dr J