Question

without actually calculating the cubes find the values of(1/2)^3+(1/3)^3-(5/6)^3​

Answer 1

Without actually calculating the cubes, we have to find,

$\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{3}\right)^3-\left(\dfrac{5}{6}\right)^3$

Here we have to recall the concept given below.

$\boxed{\begin{minipage}{11.44cm} a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac) \end{minipage}}$

But, what if  $a+b+c=0$ ???

$\begin{aligned}&a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)\\ \\ \Longrightarrow\ \ &a^3+b^3+c^3-3abc=0\\ \\ \Longrightarrow\ \ &a^3+b^3+c^3=3abc\end{aligned}$

We get this.

But why  $a+b+c=0$  is taken?!

Because, it seemed that,

$\large \displaystyle \frac{1}{2}+\frac{1}{3}-\frac{5}{6}=0$

So, on taking,

$\leadsto\ a=\dfrac{1}{2}\\ \\ \\ \leadsto\ b=\dfrac{1}{3}\\ \\ \\ \leadsto\ c=-\dfrac{5}{6}$

We absolutely get,

$\begin{aligned}&\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^3-\left(\frac{5}{6}\right)^3=3\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)\left(-\frac{5}{6}\right)\\ \\ \Longrightarrow\ \ &\left(\frac{1}{2}\right)^2+\left(\frac{1}{3}\right)^3-\left(\frac{5}{6}\right)^3=\large\text{ \bold{\ \ -\dfrac{5}{12}} }\end{aligned}$

Since  $a+b+c=0.$

So the answer is  -5/12.

Hence found the value without actually calculating the cubes!!!

Answer 2

Answer:

Step-by-step explanation:

Solution :

***************************************

We know that ,

If x + y + z = 0 , then

x³ + y³ + z³ = 3xyz

****************************************

Here ,

( 1/2 )³+( 1/3 )³ - ( 5/6 )³

= ( 1/2 )³ + ( 1/3 )³ + ( -5/6 )³

x = 1/2 , y = 1/3 , z = -5/6

Now ,

x + y + z

= 1/2 + 1/3 + ( -5/6 )

= 1/2 + 1/3 - 5/6

= ( 3 + 2 - 5 )/6 [ Since LCM = 6 ]

= 0/6

= 0

( 1/2 )³ + ( 1/3 )³ + ( -5/6 )³

= 3xyz

= 3( 1/2 )( 1/3 )( -5/6 )

= - 5/12

Therefore ,

( 1/2 )³ + ( 1/3 )³ + ( -5/6 )³

= -5/12