Question
Factorise: $a^3(b-c)^3+b^3(c-a)^3+c^3(a-b)^3$
✓
Answer
We have: $a^3(b - c)^3 + b^3(c - a)^3 + c^3(a - b)^3$
$= [a(b - c)]^3 + [a(b - c)]^3 + [b(c - a)]^3 + [c(a - b)]^3$
Put, $a(b - c) = x, b(c - a) = y, c(a - b) = z$
Here, $x + y + z = a(b - c) + b(c - a) + c(a - b)$
$= ab - ac + bc - ab - ab + ac - bc$
Thus, We have: $a^3(b - c)^3 + b^3(c - a)^3 + c^3(a - b)^3$
$= x^3 + y^3 + z^3 = 3xyz$ [When $x + y + z = 0, x^3 + y^3 + z^3 = 3xyz]$
$= 3a(b - c)b(c - a)c(a -b) = 3abc(a - b)(b - c)(c - a)$
$= [a(b - c)]^3 + [a(b - c)]^3 + [b(c - a)]^3 + [c(a - b)]^3$
Put, $a(b - c) = x, b(c - a) = y, c(a - b) = z$
Here, $x + y + z = a(b - c) + b(c - a) + c(a - b)$
$= ab - ac + bc - ab - ab + ac - bc$
Thus, We have: $a^3(b - c)^3 + b^3(c - a)^3 + c^3(a - b)^3$
$= x^3 + y^3 + z^3 = 3xyz$ [When $x + y + z = 0, x^3 + y^3 + z^3 = 3xyz]$
$= 3a(b - c)b(c - a)c(a -b) = 3abc(a - b)(b - c)(c - a)$
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
In the given figure, if $\text{AB}=\text{AC}$ and $\angle\text{B}=\angle\text{C}.$ Prove that $\text{PQ}=\text{CP}.$ 
→In the following, using the remainder theorem, find the remainder when $f(x)$ is divided by $g(x)$ and verify the by actual division: $f(x) = 4x^4 - 3x^3 - 2x^2 + x - 7, g(x) = x - 1$
→Prove that the angle in a segment shorter than a semicircle is greater than a right angle.
A total of $25 $patients admitted to a hospital are tested for levels of blood sugar, (mg/ dl) and the results obtained were as follows:
Find mean, median and mode (mg/ dl) of the above data.
→|
$87$
|
$71$
|
$83$
|
$67$
|
$85$
|
|
$77$
|
$69$
|
$76$
|
$65$
|
$85$
|
|
$85$
|
$54$
|
$70$
|
$68$
|
$80$
|
|
$73$
|
$78$
|
$68$
|
$85$
|
$73$
|
|
$81$
|
$78$
|
$81$
|
$77$
|
$75$
|
In a hot water heating system, there is a cylindrical pipe of length 28m and diameter $5\ cm$. Find the total radiating surface in the system.
→In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.
Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side if intersect, they will intersect on the circumcircle of the triangle.
In Fig. $X$ and $Y$ are the mid-points of $AC$ and $AB$ respectively, $QP || BC$ and $CYQ$ and $BXP$ are straight lines. Prove that $ar\ (ABP) = ar\ (ACQ).$
→Determine the measure of each of the equal angles of a right-angled isosceles triangle. $OR ABC$ is a right-angled triangle in which $\angle\text{A}=90^\circ$ and $\text{AB}=\text{AC}.$ Find $\angle\text{A}$ and $\angle\text{C}.$
→Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side if intersect, they will intersect on the circumcircle of the triangle.