Gujarat BoardEnglish MediumSTD 9MathsPolynomials2 Marks
Question
Without finding the cubes, factorise:
$(x-2 y)^3+(2 y-3 z)^3+(3 z-x)^3$
✓
Answer
We know that,
$a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
Also, if $a+b+c=0$, then $a^3+b^3+c^3=3 a b c$
Here, we see that $(x-2 y)+(2 y-3 z)+(3 z-x)=0$
Therefore, $(x-2 y)^3+(2 y-3 z)^3+(3 z-x)^3=3(x-2 y)(2 y-3 z)(3 z-x)$.
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