Question

Without actually calculating the cubes, find the values of \[{{\left( 28 \right)}^{3}}+{{\left( -15 \right)}^{3}}+{{\left( -13 \right)}^{3}}\].

Answer 1

Hint: Observe the relation among the numbers 28, -15 and -13. Use the algebraic identity of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$

Complete Step-by-Step solution:
To solve the problem. It is given as
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)$
Where, if a + b + c = 0 then the left hand side of the equation will become 0. Solve it by using this identity.
As we have to determine the value of \[{{\left( 28 \right)}^{3}}+{{\left( -15 \right)}^{3}}+{{\left( -13 \right)}^{3}}\] without finding the cubes of these numbers. Hence, we need to observe some relation between these numbers.
So, we observe that sum of numbers 28, -15 and -13 is 0 (28 + (-15) + (-13) = 28 – 15 – 13 = 0).
Now, we know the identity of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$ is given as
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ca \right)$…………… (i)
Now, we observe from the relation above that term ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$ will be 0 if sum of three numbers a, b, c is 0 (R.H.S.).
So, we can put the given numbers 28, -15 and -13 to the given expression such that a = 28, b = -15 and c = -13 to get an answer. It means, we get the relation from the equation (i) if a + b + c = 0 as
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=0$ if a + b + c = 0,
${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$ if a + b + c = 0 …………. (ii)
Now, put the values of a, b, c as 28, -15, -13 and hence we get
 \[{{\left( 28 \right)}^{3}}+{{\left( -15 \right)}^{3}}+{{\left( -13 \right)}^{3}}=3\times \left( 28 \right)\times \left( -15 \right)\times \left( -13 \right)\]
Where a + b + c = 28 -15 -13 = 0
So, we get
\[{{\left( 28 \right)}^{3}}+{{\left( -15 \right)}^{3}}+{{\left( -13 \right)}^{3}}=3\times 28\times \left( -15 \right)\times \left( -13 \right)\]
\[{{\left( 28 \right)}^{3}}+{{\left( -15 \right)}^{3}}+{{\left( -13 \right)}^{3}}=16380\]
Hence, the value of the given expression in the problem will be 16380.
We used the identity (i) because the term (a + b + c) in the right hand side of it will be 0 after putting a = 28, b = -15, c = -13.
So, here, we did not calculate the cubes of the individual numbers to get the value of the expression \[{{\left( 28 \right)}^{3}}+{{\left( -15 \right)}^{3}}+{{\left( -13 \right)}^{3}}\]. Hence, 16380 is the correct answer.

Note: Observing the relation 38 – 15 – 13 = 0 is the key point of question for applying the identity of ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$ .
Another approach for the question would be that we can apply the identity of ${{a}^{3}}+{{b}^{3}}$ with any two of terms and hence simplify them. Here, one needs to calculate the square of the number but cubes. So, it can be another approach. Identity of ${{a}^{3}}+{{b}^{3}}$ is given as
${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$
If a + b + c = 0 then ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc$ is very important property of the algebraic identities. There are a large number of problems in mathematics only based on it. So, just remember it for further same type of questions.