7^3+21^3
7^3 + 7^3*3^3 => 7^3(1+3^3)= 7^3*28=7^3*7*4= 7^4*2^2

answer is a

\(7^3 + 21^3\)

= \(7^3 + 7^3*3^3\)

= \(7^3 (1 +3^3 )\)

= \(7^3 *28\)

= \(7^3 *7*2^2\)

= \(7^4 *2^2\)

Hence, answer will be (A) \(7^4 *2^2\)
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This question can also be solved without using any exponent laws.

First, we'll use the fact that the product of a bunch of odd integers will always be odd.
And we'll use the fact that, if a product contains 1 or more even integers, then that product will be even

So, 7³ + 21³ = (some odd number)³ + (some odd number)³
= ODD + ODD
= EVEN
So, 7³ + 21³ = an even number

Now check the answer choices....
A. \(7^4 * 2^2\) = \(ODD^4*EVEN^2\) = ODD * EVEN = EVEN (Keep)

B. \(7^3*2^4\) = \(ODD^3*EVEN^4\) = ODD * EVEN = EVEN (Keep)

C. \(7^3*3^3\) = \(ODD^3 * ODD^3\) = ODD * ODD = ODD (eliminate)

D. \(7^3*3^4\) = \(ODD^3 * ODD^4\) = ODD * ODD = ODD (eliminate)

E. \(7^4*3^3\) = \(ODD^4 * ODD^3\) = ODD * ODD = ODD (eliminate)

So, the correct answer is A or B

At this point, let's focus on the UNITS DIGITS ONLY

\(7^3\) = (7)(7)(7) = ????3 [some number with units digit 3]
\(21^3 =\) (21)(21)(21) = ????1 [some number with units digit 1]
So, \(7^3 + 21^3 =\) ????3 + ????1 = ????4
Now we'll check the remaining answer choices to see which one yields an answer with units digit 4

A. \(7^4 * 2^2\)
\(7^4\) = (7)(7)(7)(7) = ????1 [some number with units digit 1]
\(2^2\) = (2)(2) = 4
So, \(7^4 * 2^2\) = (????1)(4) = ??????4
Great - keep A

B. \(7^3*2^4\)
\(7^3\) = (7)(7)(7) = ????3
\(2^2\) = (2)(2)(2)(2) = 16
So, \(7^3 * 2^4\) = (????3)(16) = ??????8
ELIMINATE B

Answer: A
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