(2a + 1)³

I think I have heard of binomial expansion, but that is as far as I have got with it. I think I shall save that for another day to save frying my brain too much now.

Is that other reply after yours also an answer for this? 

(2a+1)³

We could first look to combine like terms to simplify this but there are none we could combine.

Then we could look at order of operations, also known as precedence. There are a number of mneumonics to help us remember the order to do things, PEMDAS (Parenthesis, Exponents, Multiplication, Division, Addition, Subtraction), BIDMAS (Brackets, Indecies, Division, Multiplication, Addition, Subtraction) and a few more depending on where you have been taught, but thankfully they all basically mean the same thing.

This means we should do any work we can within the brackets/parenthesis first (following the order of the mneumonic) followed by any exponents/indecies (including roots), followed by multiplication/division (these can be either way around as long as they follow brackets/parenthesis and exponents/indicies) followed by addition/subtraction (these can be either way around as long as they are the last to be completed).

In this case we can't calculate the contents of the brackets/parenthesis as it contains numbers and an algebraic term. Therefor we go straight to applying the exponent/indicies which in this case is "³", meaning cubed or to the power of 3. Put simply this is equivelant to (2a+1)*(2a+1)*(2a+1) though we don't need the * (multiplication) symbol inbetween the brackets as we know that back to back brackets automatically have their contents multiplied, so would be written;

= (2a+1)(2a+1)(2a+1)

We could then multiply the contents of the first two brackets/parenthesis against each other. I use the FOIL method. First from each bracket, Outside from each bracket, Inside from each bracket, Last from each bracket. It ensures that each term from one bracket is multiplied by each term from the other bracket. The order is not important but ensuring all are completed is;

= ((2a*2a)+(2a*1)+(1*2a)+(1*1))(2a+1)

= (4a²+2a+2a+1)(2a+1)

To simplify this we could combine like terms within the brackets/parenthesis;

= (4a²+4a+1)(2a+1)

= We could now multiply the contents of the two remaining brackets/parenthesis though this time we can't use the FOIL method as we have three terms inside the first bracket/parenthesis, another methodical approach will have to be used to ensure all terms from the first bracket/parenthesis are multiplied by all terms in the second bracket/parenthesis;

= (4a²*2a)+(4a²*1)+(4a*2a)+(4a*1)+(1*2a)+(1*1)

= 8a³+4a²+8a²+4a+2a+1

We could then combine like terms leaving us with;

= 8a³+12a²+6a+1

Sorry if this is a bit long winded but I've tried to cover everything for you here. Please bear in mind I am just a newbie so there may be more elegant solutions to this and I may be wrong, though I believe I have it correct.

Thanks Tenacious   

You can certainly do it the way Tenacious did.

Her is another approach (some of you won't know what i am talking about yet)

I have used a binomial expansion.

$$\\(2a+1)^3=(2a)^3+3*(2a)^2+3*(2a)+1\\
(2a+1)^3=8a^3+3*4a^2+3*(2a)+1\\
(2a+1)^3=8a^3+12a^2+6a+1\\$$

$$\frac{\left(\sqrt[3]{x}-1\right)}{2}$$

Yes Tenacious, you are not ready for a binomial expansion just yet.  :))