MCQ
$(a-b)^3+(b-c)^3+(c-a)^3=$
- A$(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
- B$(a-b)(b-c)(c-a)$
- ✓$3(a-b)(b-c)(c-a)$
- DNone of these.
✓
Answer
Correct option: C.
$3(a-b)(b-c)(c-a)$
Let
$a-b=A$
$b-c=B$
$c-a=c$
Now $(A+B+C)^3=A^3+B^3+C^3+3(A+B)(B+C)(C+A)$
$\Rightarrow A^3+B^3+C^3=(A+B+C)^3-3(A+B)(B+C)(C+A)$
Now putting values of $A, B$ and $C$. we get
$(a-b)^3+(b-c)^3+(c-a)^3$
$=(a-\not b+\not b-\not+\not-\not a)^3 -3(a-\not b+\not b-c)(b-c t+c-a)(c-\not a+\not a-b)$
$\Rightarrow(a-b)^3+(b-c)^3+(c-a)^3$
$=0-3(a-c)(b-a)(c-b)$
$\Rightarrow(a-b)^3+(b-c)^3+(c-a)^3$
$=3(a-b)(b-c)(c-a)$
Hence, correct option is $(c).$
$a-b=A$
$b-c=B$
$c-a=c$
Now $(A+B+C)^3=A^3+B^3+C^3+3(A+B)(B+C)(C+A)$
$\Rightarrow A^3+B^3+C^3=(A+B+C)^3-3(A+B)(B+C)(C+A)$
Now putting values of $A, B$ and $C$. we get
$(a-b)^3+(b-c)^3+(c-a)^3$
$=(a-\not b+\not b-\not+\not-\not a)^3 -3(a-\not b+\not b-c)(b-c t+c-a)(c-\not a+\not a-b)$
$\Rightarrow(a-b)^3+(b-c)^3+(c-a)^3$
$=0-3(a-c)(b-a)(c-b)$
$\Rightarrow(a-b)^3+(b-c)^3+(c-a)^3$
$=3(a-b)(b-c)(c-a)$
Hence, correct option is $(c).$
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