Question
Download Solution PDFSimplify the given expression. \(\frac{(2.3)^3+(15.9)^3+(3.7)^3-3 \times 2.3 \times 15.9 \times 3.7}{(2.3)^2+(15.9)^2+(3.7)^2-2.3 \times 15.9-15.9 \times 3.7-3.7 \times 2.3}\)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
Expression: \(\dfrac{(2.3)^3+(15.9)^3+(3.7)^3-3 \times 2.3 \times 15.9 \times 3.7}{(2.3)^2+(15.9)^2+(3.7)^2-2.3 \times 15.9-15.9 \times 3.7-3.7 \times 2.3}\)
Formula used:
For any real numbers a, b, and c:
a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)
Calculation:
Given expression: \(\dfrac{(2.3)^3+(15.9)^3+(3.7)^3-3 \times 2.3 \times 15.9 \times 3.7}{(2.3)^2+(15.9)^2+(3.7)^2-2.3 \times 15.9-15.9 \times 3.7-3.7 \times 2.3}\)
Using the formula:
⇒ \(\dfrac{(2.3 + 15.9 + 3.7)(2.3^2 + 15.9^2 + 3.7^2 - 2.3 \times 15.9 - 15.9 \times 3.7 - 3.7 \times 2.3)}{(2.3^2 + 15.9^2 + 3.7^2 - 2.3 \times 15.9 - 15.9 \times 3.7 - 3.7 \times 2.3)}\)
⇒ \(\dfrac{(2.3 + 15.9 + 3.7)}{1}\)
⇒ 2.3 + 15.9 + 3.7
⇒ 21.9
∴ The correct answer is option (4).
Last updated on Jun 3, 2025
-> The Staff Selection Commission has officially released the SSC Selection Post Phase 13 Notification 2025 on its official website at ssc.gov.in.
-> A total number of 2423 Vacancies have been announced for various selection posts under Government of India.
-> The Computer Based Exam is scheduled from 24th July to 4th August, 2025. Candidates will be able to apply online from 2nd June 2025 to 23rd June 2025.
-> The SSC Selection Post Phase 13 exam is conducted for recruitment to posts of Matriculation, Higher Secondary, and Graduate Levels.
-> The selection process includes a CBT and Document Verification.
-> Some of the posts offered through this exam include Laboratory Assistant, Deputy Ranger, Upper Division Clerk (UDC), and more.
-> Enhance your exam preparation with the SSC Selection Post Previous Year Papers & SSC Selection Post Mock Tests for practice & revision.