How do you find the sum of cube ?
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Answer:
[tex](2x+5)(4x^{2} -10x+25)=0[/tex]
Step-by-step explanation:
Given:
The given equation is.
[tex]8x^{3} +125=0[/tex]
Find the some of cube.
Solution:
[tex]8x^{3} +125=0[/tex]
[tex]2^{3}x^{3} +5^{3}=0[/tex]
[tex](2x)^{3} +5^{3}=0[/tex]----------(1)
The sum of the cube formula is given below.
[tex](a^{3} +b^{3})=(a+b)(a^{2} -ab+b^{2} )[/tex]-----------(2)
By comparing equation 1 and equation 2
[tex]a=2x, b=5[/tex]
substitute a and b value in equation 2
[tex]((2x)^{3} +5^{3})=(2x+5)((2x)^{2} -(2x)(5)+5^{2})[/tex]
[tex]((2x)^{3} +5^{3})=(2x+5)(4x^{2} -(10x)+25)[/tex]
[tex]((2x)^{3} +5^{3})=(2x+5)(4x^{2} -10x+25)[/tex]
Therefore the sum of the cube [tex](2x+5)(4x^{2} -10x+25)=0[/tex]