Without actually calculating the value of cubes find the value of $(1)^3+(2)^3+2(4)^3+(-5)^3+(-6)^3$. Also write the identity used

Question

Without actually calculating the value of cubes find the value of $(1)^3+(2)^3+2(4)^3+(-5)^3+(-6)^3$. Also write the identity used

Answer 1

You can use the identity that if $a + b + c=0$, then $a^3 + b^3 + c^3 = 3abc$. So we can say that:

• $2^3 + 4^3 + (-6)^3 = 3\cdot 2\cdot 4\cdot -6 = -144$.

• $1^3 + 4^3 + (-5)^3 = 3\cdot 1\cdot 4 \cdot -5 = -60$.

Adding both, we get:

$1^3 + 2^3 + 2 (4)^3 + (-5)^3 + (-6)^3 = -204$

Answer 2

Let $S=1^3+2^3+4^3-5^3-6^3.$ Then $$S=3(1^3+2^3+3^3+4^3)-2(1^3+2^3+3^3)\\ -(1^3+2^3+3^3+4^3+5^3+6^3)+(1^3+2^3).$$ Now apply the formula $1^3+2^3+\cdots+k^3=(1+2+\cdots +k)^2.$ Then $$S=3\cdot 10^2-2\cdot 6^2 - 21^2 +3^2=-204.$$ This may not be what you want since we still need to square numbers to finish. [At least we don't have to cube any of them :)] But it seems there must be some computation necessary to get the result.

Answer 3

use the sum of cubes of continuous integers: $\dfrac{n^2 (n+1)^2}{ 4}$ , this is the same as $1^3 + 2^3 +\cdots+n^3$. You can use this. But you need to have coefficients of 1 in front of your cubes, so split the $(2)4^3=4^3+4^3$. Leave the negatives at the end and do the same thing