Question : If $a+b-c=20$ and $a^2+b^2+c^2=152$, find the value of $a^3+b^3-c^3+3abc$.
Option 1: 480
Option 2: 720
Option 3: 640
Option 4: 560
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 560
Solution : Given: $a+b-c=20$ and $a^2+b^2+c^2=152$ $ a+b-c = 20$ Squaring both sides, we get, ⇒ $a^2 + b^2+c^2 + 2ab-2bc-2ca = 400$ ⇒ $2(ab-bc-ca) = 400-152$ ⇒ $ab-bc-ca = 124$ Now to find, $a^3+b^3-c^3+3abc$ $= (a+b-c)(a^2+b^2+c^2-(ab-bc-ca))$ $= 20\times (152-124)$ $= 20\times 28 = 560$ Hence, the correct answer is 560.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $(a+b+c)=20$ and $a^2+b^2+c^2=152$, find the value of $a^3+b^3+c^3-3 abc$.
Question : If $(a+b+c)=14$ and $\left(a^3+b^3+c^3-3abc\right)=98$, find the value of $\left(a^2+b^2+c^2\right)$.
Question : What is the value of ${a}^3+{b}^3+{c}^3$ if $(a+b+c)=0$?
Question : If (a + b + c) = 12 and (ab + bc + ca) = 47, find the value of (a3 + b3 + c3 – 3abc).
Question : If a + b + c = 0, the value of (a3 + b3 + c3) is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile