Evaluate:
(a) $25^{3}-75^{3}+50^{3}$
(b) $48^{3}-30^{3}-18^{3}$
(c) $(1 / 2)^{3}+(1 / 3)^{3}-(5 / 6)^{3}$
(d) $(0.2)^{3}-(0.3)^{3}+(0.1)^{3}$
Given,
(a) $25^{3}-75^{3}+50^{3}$
we know that,
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$ here,
$a=25, b=-75, c=50$
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$a^{3}+b^{3}+c^{3}=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=(25-75+50)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=(0)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=3 a b c$
$25^{3}+(-75)^{3}+50^{3}=3 a b c$
$=3(25)(-75)(50)$
$=-281250$
Hence, the value $25^{3}+(-75)^{3}+50^{3}=-281250$
(b) $48^{3}-30^{3}-18^{3}$
we know that,
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
here, $a=48, b=-30, c=-18$
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$a^{3}+b^{3}+c^{3}=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=(48-30-18)$
$\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=(0)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=3 a b c$
$48^{3}+(-30)^{3}+(-18)^{3}=3 a b c$
$=3(48)(-30)(-18)=77760$
(c) $(1 / 2)^{3}+(1 / 3)^{3}-(5 / 6)^{3}$
we know that,
$a^{3}+b^{3}+c^{3}-3 a b c$
$=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$ here,
$a=1 / 2, b=1 / 3, c=-5 / 6$
$a^{3}+b^{3}+c^{3}-3 a b c$
$=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$
$a^{3}+b^{3}+c^{3}=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=(1 / 2+1 / 3-5 / 6)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c b y$
Using least common multiple
$\mathrm{a}^{3}+\mathrm{b}^{3}+\mathrm{c}^{3}=\left(\frac{1 * 6}{2 * 6}+\frac{1 * 4}{3 * 4}-\frac{5 * 2}{6 * 2}\right)$
$\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=(6 / 12+4 / 12-10 / 12)$
$\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=0$
$\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=3 a b c$
$(1 / 2)^{3}+(1 / 3)^{3}-(-5 / 6)^{3}=3 \times 1 / 2 \times 1 / 3 \times-5 / 6$
$=1 / 2 *-5 / 6=-5 / 12$
Hence, the value of
$(1 / 2)^{3}+(1 / 3)^{3}-(5 / 6)^{3}$ is $-5 / 12$
(d) $(0.2)^{3}-(0.3)^{3}+(0.1)^{3}$
we know that, $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)$
$\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$ here,
$a=0.2, b=0.3, c=0.1$
$a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)$
$\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) a^{3}+b^{3}+c^{3}=(a+b+c)$
$\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=(0.2-0.3+0.1)$
$\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=(0)$
$\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)+3 a b c$
$a^{3}+b^{3}+c^{3}=3 a b c$
$(0.2)^{3}-(0.3)^{3}+(0.1)^{3}=3 a b c$
$=3(0.2)(-0.3)(0.1)=-0.018$
Hence, the value $(0.2)^{3}-(0.3)^{3}+(0.1)^{3}$ is $0.018$
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