Manipulating exponents: not sure how this example is worked out

[geologist]

Summary: An example problem on Brilliant.org leaves out a step in solving 2^3 + 6^3

I've spend a couple days trying to understand how the second step of solving the problem 2³ + 6³, which is 2³ + 6³ = 2³ + (2³ * 3³) = 2³(1 + 3³).

It is not clear how this problem gets to 2³(1 + 3³). I can see that somehow you have to factor one of the 2³, but I cannot visualize how that is done. What concept should I review to understand this problem?

 

[symbolipoint]

Just a quick look at your post #1,
...
DISTRIBUTIVE PROPERTY

 

[symbolipoint]

2^3+6^3
2^3+(2*3)^3
2^3+(2^3*3^3)
2^3+(2^3)(3^3)
(2^3)(1+3^3), factorizable because of the Distributive Property

 

[fresh_42]

I cannot visualize how that is done.

It is how an area can be calculated:

 

[symbolipoint]

fresh_42, I have occasionally used that kind of visualizing, too. When trying to factorize an expression, one must actually see the factors IN the expression.

 

[geologist]

2^3+6^3
2^3+(2*3)^3
2^3+(2^3*3^3)
2^3+(2^3)(3^3)
(2^3)(1+3^3), factorizable because of the Distributive Property

I understand how a(b-c) = ab - ac and how (5x+2)(5x+2) = 25x^2 + 20x + 4, but I'm struggling to understand what is "a" in the above problem. I'll have to digest this information and read more examples on the distributive property.

 

[fresh_42]

I understand how a(b-c) = ab - ac and how (5x+2)(5x+2) = 25x^2 + 20x + 4, but I'm struggling to understand what is "a" in the above problem. I'll have to digest this information and read more examples on the distributive property.

The expression you have is ##a+(a\cdot b)##. Now we first write it as ##a\cdot 1 + a\cdot b## so we have a common factor ##a## in both terms, which allows us to use the distributive law backwards:
$$a\cdot 1 + a\cdot b = a\cdot (1+b)$$

 

[geologist]

The expression you have is ##a+(a\cdot b)##. Now we first write it as ##a\cdot 1 + a\cdot b## so we have a common factor ##a## in both terms, which allows us to use the distributive law backwards:
$$a\cdot 1 + a\cdot b = a\cdot (1+b)$$

That make sense. I understand what’s going on now. The step 2^3+(2^3*3^3) is unnecessary. My confusion with 2^3(1+3^3) can be understood better looking at the originally stated problem 2^3 + 3^3. Now it’s easy to see how 2^3 factors out.

 

[symbolipoint]

That make sense. I understand what’s going on now. The step 2^3+(2^3*3^3) is unnecessary. My confusion with 2^3(1+3^3) can be understood better looking at the originally stated problem 2^3 + 3^3. Now it’s easy to see how 2^3 factors out.

I was trying to show 'all' the steps to make the solution clear. Maybe just a few minutes of rest, and returning to it , the steps will seem much more understandable than before. The way I showed association might have unsettled you when first looking.