Without actually calculating the cubes,find t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $a = 0.2$,$b = -0.3$,and $c = 0.1$.We observe that $a + b + c = 0.2 + (-0.3) + 0.1 = 0.3 - 0.3 = 0$.We know the algebraic identity: If $a + b

Vedclass · Accepted Answer

$-0.018$

Vedclass · Answer

(C) Let $a = 0.2$,$b = -0.3$,and $c = 0.1$.We observe that $a + b + c = 0.2 + (-0.3) + 0.1 = 0.3 - 0.3 = 0$.We know the algebraic identity: If $a + b + c = 0$,then $a^{3} + b^{3} + c^{3} = 3abc$.Substituting the values:$(0.2)^{3} + (-0.3)^{3} + (0.1)^{3} = 3(0.2)(-0.3)(0.1)$.Calculating the product: $3 	imes 0.2 = 0.6$; $0.6 	imes (-0.3) = -0.18$; $-0.18 	imes 0.1 = -0.018$.Thus,$(0.2)^{3} - (0.3)^{3} + (0.1)^{3} = -0.018$.

Without actually calculating the cubes,find the value of each of the following:
$(0.2)^{3} - (0.3)^{3} + (0.1)^{3}$

Similar Questions

Factorise:
$3x^{3}-x^{2}-3x+1$

Factorise $x^{3}+12 x^{2}+39 x+28$.

Find the value of the following polynomial at the indicated value of the variable:
$p(x) = x^{2} + 5x - 24$ at $x = 3$

Factorise:
$2 \sqrt{2} a^{3} + 8 b^{3} - 27 c^{3} + 18 \sqrt{2} a b c$

Write the following cube in expanded form:
$(4x - 3y)^{3}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Without actually calculating the cubes,find the value of each of the following:$(0.2)^{3} - (0.3)^{3} + (0.1)^{3}$