Question

6. Write the following cubes in expanded form:

Answer 1

Answer:

Step-by-step explanation:

we can use this algebraic identity:

$(a+b)^{3} =a^{3} +b^{3} +3ab(a+b)$

So,$(3a+4b)^{3} =(3a)^{3} +(4b)^{3}+3(3a)(4b)(3a+4b)$

=$(3a+4b)^{3} =27a^{3} +64b^{3} +36ab(3a+4b)$

=$(3a+4b)^{3} =27a^{3} +64b^{3} +108a^{2} b+144ab^{2}$

$(999)^{3}$=$(1000-1)^{3}$

Now, we can use the following identity:

$(a-b)^{3} =a^{3} -b^{3} -3ab(a-b)$

=$(1000-1)^{3} =1000^{3} -1^{3} -3(1000)(1)(1000-1)$

$(1000-1)^{3} =1,000,000,000-1-2,997,000$

$(1000-1)^{3} =997,002,999$