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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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

21^3 = (7*3)^3 = 7^3*3^3

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

thus 7^3(7*2^2)
= 7^4*2^2

Ans 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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We can simplify the given equation:

7^3 + 21^3

7^3 + (7x3)^3

7^3 + 7^3 x 3^3

7^3(1 + 3^3)

7^3(1 + 27)

7^3(28)

7^3 x 7^1 x 2^2

7^4 x 2^2

Answer: A
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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

ans - A

thanks->Kudos!

Method #1: Let us calculate approximate values:

\(7^3 + 21^3 = 7^2 * 7 + 7^3 * 3^3\)

\(= 49 * 7 + 7^3 * 3^3\)

\(≈ 50 * 7 + 7^3 * 3^3\)

\(= 350 + 350 * 27\)

\(≈ 350 + 350 * 30\)

\(= 350 + 10500 = 10850\)
Note: The actual answer is less than 10850

Working with options:

A. \(7^4*2^2 = 49*49*4 ≈ 50*50*4 = 10000\) -- may be possible

B. \(7^3*2^4 = 49*7*16 ≈ 50*7*16 = 50*16*7 = 800 * 7 = 5600\) - too small

C. \(7^3*3^3 = 49*7*27 ≈ 50*7*30 = 350 * 30 = 10500\) -- may be possible

D. \(7^3*3^4 = 49 * 7 * 81 ≈ 50*7*80 = 28000\) -- too large

E. \(7^4*3^3 = 49*49*27 ≈50*50*30 = 75000\) -- too large

Thus, either A or C is possible.

Now, the given question: \(7^3 + 21^3\) is a sum of 2 odd numbers and hence, the result is even

Verifying options A and C:

A: \(7^4*2^2\) is the product of an odd number and an even number => hence, is even --- matches

C: \(7^3*3^3\) is the product of 2 odd numbers => hence, is odd --- does not match

Answer A



Alternate approach:

Work with the units digits [Note: The cycle for 3: 3,9,7,1; for 7: 7,9,3,1; 2: 2,4,8,6]:

Units digit of the question: \(7^3 + 21^3\) = Units digit of \(3 + 1 = 4\)

Working with the options:

A. Units digit of \(7^4*2^2\) = Units digit of \(1 * 4 = 4\) --- matches

B. Units digit of \(7^3*2^4\) = Units digit of \(3 * 6 = 8\) --- does not match

C. Units digit of \(7^3*3^3\) = Units digit of \(3 * 7 = 1\) --- does not match

D. Units digit of \(7^3*3^4\) = Units digit of \(3 * 1 = 3\) --- does not match

E. Units digit of \(7^4*3^3\) = Units digit of \(1 * 7 = 7\) --- does not match

Answer A



Approach #3: Those who are good with simplification, the best way of solving is as follows:

\(7^3 + 21^3 = 7^3 + 7^3 * 3^3\)

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

\(= 7^3 * (1 + 27)\)

\(= 7^3 * 28\)

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

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

Answer A
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\(7^3+21^3 = 7^3(1+3^3) = 7^3*28 = 7^4*2^2\)

IMO A

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