Question
Write the value of $25^3 - 75^3+ 50^3.$
✓
Answer
The given expression is $25^3-75^3+50^3$ Let $a =25, b=-75$ and $c =50$.
Then the given expression becomes $25^3-75^3+50^3= a ^3+ b ^3+ c ^3$
Note that $a+b+c=25+(-75)+50=25-75+50=0$
Recall the formula $a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
when $a+b+c=0$, this becomes $a^3+b^3+c^3-3 a b c$
$=0 \cdot\left(a^2+b^2+c^2-a b-b c-c a\right)=0 a^3+b^3+c^3=3 a b c$ So,
we have the new formula $a^3+b^3+c^3=3 a b c$,
when $a+b+c=0$ Using the above formula,
the value of the given expression is $a ^3+ b ^3+ c ^3=3 abc 25^3-75^3+50^3$
$=3 \cdot(25) \cdot(-75) \cdot(50) 25^3-75^3+50^3=-281250$
Then the given expression becomes $25^3-75^3+50^3= a ^3+ b ^3+ c ^3$
Note that $a+b+c=25+(-75)+50=25-75+50=0$
Recall the formula $a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
when $a+b+c=0$, this becomes $a^3+b^3+c^3-3 a b c$
$=0 \cdot\left(a^2+b^2+c^2-a b-b c-c a\right)=0 a^3+b^3+c^3=3 a b c$ So,
we have the new formula $a^3+b^3+c^3=3 a b c$,
when $a+b+c=0$ Using the above formula,
the value of the given expression is $a ^3+ b ^3+ c ^3=3 abc 25^3-75^3+50^3$
$=3 \cdot(25) \cdot(-75) \cdot(50) 25^3-75^3+50^3=-281250$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Simplify the following expressions::
$(\sqrt{3}+\sqrt{7})^2$
→$(\sqrt{3}+\sqrt{7})^2$
Write: The cofficient of $x$ in $\frac{3}{8}\text{x}^2-\frac{2}{7}\text{x}+\frac{1}{6}.$
→Find the value of $\frac{21\sqrt{12}}{10\sqrt{27}}.$
→Write the equation of the line that is parallel to $y-$axis and passing through the point: $(-4, -3)$
→Evaluate: $\big(25\big)^{\frac{3}{2}}$
→Write an equation in two variables: $x = –5$
→Line $l$ is the bisector of an angle $\angle A$ and $B$ is any point on $l. BP$ and $BQ$ are perpendicular from $B$ to the arms of $\angle A.$
Show that:
$i. \triangle APB \cong \triangle AQB$
$ii. BP = BQ$ or $B$ is equidistant from the arms of $\angle A.$
→Show that:
$i. \triangle APB \cong \triangle AQB$
$ii. BP = BQ$ or $B$ is equidistant from the arms of $\angle A.$
Find an irrational number between $5$ and $6.$
→Find the surface area of a sphere of radius: $5.6\ cm$
→Find the volume of the right circular cone with radius $3.5$ and height $12\ cm$
→