Answer:
Step-by-step explanation:
we can use this algebraic identity:
[tex](a+b)^{3} =a^{3} +b^{3} +3ab(a+b)[/tex]
So,[tex](3a+4b)^{3} =(3a)^{3} +(4b)^{3}+3(3a)(4b)(3a+4b)[/tex]
=[tex](3a+4b)^{3} =27a^{3} +64b^{3} +36ab(3a+4b)[/tex]
=[tex](3a+4b)^{3} =27a^{3} +64b^{3} +108a^{2} b+144ab^{2}[/tex]
[tex](999)^{3}[/tex]=[tex](1000-1)^{3}[/tex]
Now, we can use the following identity:
[tex](a-b)^{3} =a^{3} -b^{3} -3ab(a-b)[/tex]
=[tex](1000-1)^{3} =1000^{3} -1^{3} -3(1000)(1)(1000-1)\\[/tex]
[tex](1000-1)^{3} =1,000,000,000-1-2,997,000[/tex]
[tex](1000-1)^{3} =997,002,999[/tex]
