Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Without actually calculating the cubes, find the value of each of the following:
${{(28)}^{3}}+{{(-15)}^{3}}+{{(-13)}^{3}}$
A. $17530$
B. $14850$
C. $12022$
D. $16380$

seo-qna
Last updated date: 24th Jul 2024
Total views: 452.7k
Views today: 9.52k
Answer
VerifiedVerified
452.7k+ views
Hint: We know the formula ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz$. So here, take $x=28$, $y=-15$ and $z=-13$ and substitute in the above formula. You will get the answer. Try it.

Complete step-by-step answer:
Exponentiation is a mathematical operation, written as ${{b}^{n}}$, involving two numbers, the base $b$ and the exponent or power $n$. When n is a positive integer, exponentiation corresponds to repeated multiplication of the base: that is, ${{b}^{n}}$ is the product of multiplying in bases.
The power (or exponent) of a number says how many times to use the number in a multiplication.

It is written as a small number to the right and above the base number.
A convenient way to note that a number is to be multiplied by itself and then itself again (for example $5\times 5\times 5$) is to say that the number is cubed. To help visualize this, picture a cube which is $5$ units long, $5$ units wide, and $5$ units high. This cube will then have a volume of $5\times 5\times 5$ or $125$ cubic units. A nice way to write "$5$ cubed" is ${{5}^{3}}$where $5$ is called the base and $3$ is called the exponent.

In arithmetic and algebra, the cube of a number $n$ is its third power: the result of the number multiplied by itself twice:
${{n}^{3}}=n\times n\times n$
It is also the number multiplied by its square:
${{n}^{3}}=n\times {{n}^{2}}$
This is also the volume formula for a geometric cube with sides of length n, giving rise to the name. The inverse operation of finding a number whose cube is n is called extracting the cube root of n. It determines the side of the cube of a given volume. It is also raised to the one-third power.
Both cube and cube root are odd functions:
${{(-n)}^{3}}=-({{n}^{3}})$

Now we know the formula ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz$.

Now let $x=28$, $y=-15$ and $z=-13$.
Substituting the values we get,

${{(28)}^{3}}+{{(-15)}^{3}}+{{(-13)}^{3}}=(28-15-13)({{(28)}^{2}}+{{(-15)}^{2}}+{{(-13)}^{2}}-(28)(-15)-(-15)(-13)-(-13)(28))+3(28)(-15)(-13)$
So we can see in above that $(28-15-13)=0$.
So we get,

$\begin{align}
& {{(28)}^{3}}+{{(-15)}^{3}}+{{(-13)}^{3}}=0+3(28)(-15)(-13) \\
& {{(28)}^{3}}+{{(-15)}^{3}}+{{(-13)}^{3}}=3(28)(-15)(-13)=16380 \\
\end{align}$

${{(28)}^{3}}+{{(-15)}^{3}}+{{(-13)}^{3}}=16380$
Here we get the value ${{(28)}^{3}}+{{(-15)}^{3}}+{{(-13)}^{3}}=16380$.

Note: Read the question carefully. Do not make silly mistakes. Also, don’t jumble while simplifying. Solve it step by step. Don’t miss any term. Take utmost care that you are going step by step way. Your concepts regarding powers and exponents should be cleared. Also, you must know the formula
${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=(x+y+z)({{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy-yz-zx)+3xyz$.