Find the value of $(2)^{3}+(3)^{3}+(-5)^{3}$

Question

Accepted Answer

The correct option is 2:
To solve this problem, we need to find the cube of each number and then add them together.
Recall that the cube of a number $a$ is given by $a^3 = a 	imes a 	imes a$.

Step 1: Calculate the cube of each number in the expression.
For $(2)^3$:
$(2)^3 = 2 	imes 2 	imes 2 = 8$

For $(3)^3$:
$(3)^3 = 3 	imes 3 	imes 3 = 27$

For $(-5)^3$:
$(-5)^3 = (-5) 	imes (-5) 	imes (-5) = (25) 	imes (-5) = -125$

Step 2: Add the calculated cube values.
Now, substitute these values back into the original expression:
$(2)^{3}+(3)^{3}+(-5)^{3} = 8 + 27 + (-125)$
First, add the positive numbers:
$8 + 27 = 35$
Now, add this sum to the negative number:
$35 + (-125) = 35 - 125 = -90$

So, the value of $(2)^{3}+(3)^{3}+(-5)^{3}$ is $-90$.

Why Option 2 is correct:
Option 2 has the value $-90$, which matches our calculated result. This demonstrates a correct understanding and application of cubing positive and negative numbers and then performing addition/subtraction.

The incorrect option is 1:
Option 1 is $0$. This would only be correct if the sum of the cubes cancelled out to zero, which is not the case here. A common mistake could be thinking that if $a+b+c=0$, then $a^3+b^3+c^3=3abc$, and if $3abc=0$, then $a^3+b^3+c^3=0$. However, in this case, $2+3+(-5) = 0$, so $2^3+3^3+(-5)^3 = 3 	imes 2 	imes 3 	imes (-5) = 18 	imes (-5) = -90$. The value $0$ is not the result.

The incorrect option is 3:
Option 3 is $1$. This value is significantly different from the correct answer and might result from incorrect calculations of the individual cubes or errors in addition/subtraction. For example, a common error could be miscalculating $(-5)^3$ as $125$ instead of $-125$. If it were $8 + 27 + 125 = 160$, which is still not $1$.

The incorrect option is 4:
Option 4 is $30$. This could be a result of several calculation errors, such as incorrectly evaluating the cubes or making arithmetic mistakes during the sum. For instance, if one only considered the magnitude of numbers and performed addition without considering the signs correctly, it might lead to a positive incorrect result.