Answer:
![(2x+5)(4x^(2) -10x+25)=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/6w83vl7lmutvjkfq8p5w7b6wusy2tbandh.png)
Explanation:
Given:
The given equation is.
![8x^(3) +125=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/u4zm7wypzqbrhxbin75rt9a5lt3lva6dfg.png)
Find the some of cube.
Solution:
![8x^(3) +125=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/u4zm7wypzqbrhxbin75rt9a5lt3lva6dfg.png)
![2^(3)x^(3) +5^(3)=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/qeje77cuxo4l8n22hupjjstm8cztc7zyia.png)
----------(1)
The sum of the cube formula is given below.
-----------(2)
By comparing equation 1 and equation 2
![a=2x, b=5](https://img.qammunity.org/2020/formulas/mathematics/middle-school/l7jmv5okobzwoxhw6p1v8opunm7rpyil6n.png)
substitute a and b value in equation 2
![((2x)^(3) +5^(3))=(2x+5)((2x)^(2) -(2x)(5)+5^(2))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/dixa55yj104od5zwpexewv1ncoj8xzwc12.png)
![((2x)^(3) +5^(3))=(2x+5)(4x^(2) -(10x)+25)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/pqhr3a86w00ivirxhw4p4dar4vxvib7mva.png)
![((2x)^(3) +5^(3))=(2x+5)(4x^(2) -10x+25)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/b0m10u14mf4rsz2jimt8a5hofuehqdhe6x.png)
Therefore the sum of the cube
![(2x+5)(4x^(2) -10x+25)=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/6w83vl7lmutvjkfq8p5w7b6wusy2tbandh.png)