Question

Question 37
Without actually calculating the cubes, find the value of
(i) ${\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{3}\right)}^3-{\left(\frac{5}{6}\right)}^3$

(ii) $(0.2{)}^3-(0.3{)}^3+(0.1{)}^3$

Answer 1

(i) Given ${\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{3}\right)}^3-{\left(\frac{5}{6}\right)}^{3}or{\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{3}\right)}^3+{(-\frac{5}{6})}^3$
Here, we see that,
a=$\frac{1}{2}$,b=$\frac{1}{3}$,c=$\frac{-5}{6}$
$\frac{1}{2}+\frac{1}{3}-\frac{5}{6}=\frac{3+2-5}{6}=\frac{5-5}{6}=0$
$\therefore {\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{3}\right)}^3-{\left(\frac{5}{6}\right)}^3=3\times \frac{1}{2}\times \frac{1}{3}\times (-\frac{5}{6})=\frac{-5}{12}$
$[Usingtheidentity,ifa+b+c=0,then{a}^3+b^3+c^3=3abc]$


(ii) Given,
$(0.2{)}^3+(-0.3{)}^3+(0.1{)}^3$
Here, we see that,
a=0.2,b=-0.3,c=0.1
$0.2–0.3+0.1=0.3–0.3=0$
$\therefore (0.2{)}^3+(-0.3{)}^3+(0.1{)}^3=3\times (0.2)\times (-0.3)\times (0.1)$
$[Usingtheidentity,ifa+b+c=0,then{a}^3+b^3+c^3=3abc]$

$=-0.6\times 0.03=-0.018$