Question

without actually calculating the cubes find the value of 48 cube minus 30 cube minus 18 cube​

Answer 1

Answer:

77760

Step-by-step explanation:

48-18-30:00

then A3+b3+C3=3abc

Answer 2

Here we may recall the identity,

$(a+b+c)(a^2+b^2+c^2-ab-bc-ac)=a^3+b^3+c^3-3abc$

Given question is to find the value of  48³ - 30³ - 18³  without actually calculating their cubes.

In the identity we took, let,

$a^3+b^3+c^3=48^3-30^3-18^3$

From this, we get,

$a^3=48^3\ \ \ \ \ \Longrightarrow\ \ \ a=48\\ \\ b^3=-30^3\ \ \ \Longrightarrow\ \ \ b=-30\\ \\ c^3=-18^3\ \ \ \Longrightarrow\ \ \ c=-18$

First we calculate the value of  $a+b+c.$

$a+b+c=48-30-18=0$

Here it's 0.

So, consider the identity we used again.

$(a+b+c)(a^2+b^2+c^2-ab-bc-ac)=a^3+b^3+c^3-3abc$

If  $a+b+c=0,$  then the whole LHS will become 0. So,

$a^3+b^3+c^3-3abc=0$

This implies,

$a^3+b^3+c^3=3abc$

Thus, we can say that,

"The sum of cubes of three numbers will be equal to thrice their product if and only if their sum is 0."

According to this theorem, since  $48-30-18=0,$

$48^3-30^3-18^3=3\cdot 48\cdot (-30)\cdot (-18)\\ \\ 48^3-30^3-18^3=\textbf{77760}$

Hence found the answer without actually calculating the cubes!

The answer is 77760.