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antiseptic1488 [7]
1 year ago
13

Can you see an easy way to work out the value of

3%20%2B%204%5E3%20%2B%205%5E3" id="TexFormula1" title="\sqrt{ 1^3 + 2^3 + 3^3 + 4^3 + 5^3" alt="\sqrt{ 1^3 + 2^3 + 3^3 + 4^3 + 5^3" align="absmiddle" class="latex-formula"> ? if so, describe it.
Mathematics
1 answer:
Lina20 [59]1 year ago
5 0

We have the sum of cubes identity

a^3 + b^3 = (a + b) (a^2 - ab + b^2)

and observing that 1 + 4 = 2 + 3, we have

1^3 + 4^3 = (1 + 4) (1^2 - 4 + 4^2) = 5 \times 13

and

2^3 + 3^3 = (2 + 3) (2^2 - 6 + 3^2) = 5 \times 7

Then

\sqrt{1^3+2^3+3^3+4^3+5^3} = \sqrt{5\times13 + 5\times7 + 5\times5^2} = \sqrt{5 \times (20 + 25)} \\\\ ~~~~~~~~ = \sqrt{5^2 \times (4 + 5)} = \sqrt{5^2\times9} = \sqrt{5^2\times3^2} = 5\times3 = \boxed{15}

Alternatively, we have the well-known sum of cubes formula

\displaystyle \sum_{i=1}^n i^3 = \frac{n^2(n+1)^2}4

The sum under the square root is this sum with n=5. Then

1^3+2^3+3^3+4^3+5^3 = \dfrac{5^2\times6^2}4 = 225 = 15^2

and so the square root again reduces to 15.

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