MCQ
$(\mathrm{a}-\mathrm{b})^3+(\mathrm{b}-\mathrm{c})^3+(\mathrm{c}-\mathrm{a})^3=$
- 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
$(\text{a} - \text{b})^3 + (\text{b} - \text{c})^3 + (\text{c} - \text{a})^3\\=(\not\text{a}-\not\text{b}+\not\text{b}-\not\text{c}+\not\text{c}-\not\text{a})^3\\-3(\text{a}-\not\text{b}+\not\text{b}-\text{c})(\text{b}-\not\text{c}+\not\text{c}-\text{a})(\text{c}-\not\text{a}+\not\text{a}-\text{b})$
$\Rightarrow(a-b)^3+(b-c)^3+(c-a)^3=0-3(a-c)(b-a)(c-b)$
$\Rightarrow(a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)$
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
$(\text{a} - \text{b})^3 + (\text{b} - \text{c})^3 + (\text{c} - \text{a})^3\\=(\not\text{a}-\not\text{b}+\not\text{b}-\not\text{c}+\not\text{c}-\not\text{a})^3\\-3(\text{a}-\not\text{b}+\not\text{b}-\text{c})(\text{b}-\not\text{c}+\not\text{c}-\text{a})(\text{c}-\not\text{a}+\not\text{a}-\text{b})$
$\Rightarrow(a-b)^3+(b-c)^3+(c-a)^3=0-3(a-c)(b-a)(c-b)$
$\Rightarrow(a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)$
Hence, correct option is $(c)$.
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