Question : If $(a+b+c) \neq 0$, then $(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$ is equal to:
Option 1: $a^3+b^3-c^3-3abc$
Option 2: $a^3-b^3+c^3-3abc$
Option 3: $a^3+b^3+c^3-3abc$
Option 4: $a^3+b^3+c^3+3abc$
New: SSC CHSL tier 1 answer key 2024 out | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $a^3+b^3+c^3-3abc$
Solution : Given: $(a+b+c)\left(a^2+b^2+c^2-ab-bc-ca\right)$ Simplifying this expression, we have: $(a^3+ab^2+ac^2–a^2b–abc–ca^2+a^2b+b^3+bc^2–ab^2–b^2c–abc+a^2c+b^2c+c^3–abc–bc^2–c^2a)$ = $(a^3+b^3+c^3–3abc)$ Hence, the correct answer is $(a^3+b^3+c^3–3abc)$.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : If $\left (2a-1 \right )^{2}+\left (4b-3 \right)^{2}+\left (4c+5 \right)^{2}=0$, then the value of $\frac{a^{3}+b^{3}+c^{3}-3abc}{a^{2}+b^{2}+c^{2}}$ is:
Question : If $a^{2}+b^{2}+c^{2}-ab-bc-ca=0$, then
Question : If $\small x=a\left (b-c \right),\; y=b\left (c-a \right) ,\; z=c\left (a-b \right)$, then the value of $\left (\frac{x}{a} \right)^{3}+\left (\frac{y}{b} \right)^{3}+\left (\frac{z}{c} \right)^{3}$ is:
Question : If $a+b :\sqrt{ab} = 4:1 $ where $ a > b > 0$, then $ a:b$ is:
Question : If $\left (a+b \right):\left (b+c \right):\left (c+a \right)= 6:7:8$ and $\left (a+b+c \right) = 14,$ then value of $c$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile