ABC26GN1224 · Simplification

Subject: General Aptitude · Chapter: Simplification · Exam: · Marks: · Difficulty:

$(32)^{3}+(79)^{3}-(111)^{3}+3 \times 32 \times 79 \times 111$ is equal to
(a)0
(b)1
(c)10000
(d)30007
Answer
Answer (as printed): A
Explanation
Let $a=32, b=79, c=-111$. Then, $a+b+c=0$. So, $a^{3}+b^{3}+c^{3}=3 a b c \Rightarrow(32)^{3}+(79)^{3}-(111)^{3}$ $$=3 \times 32 \times 79 \times(-111)$$ $\therefore \quad$ Given exp. $=-(3 \times 32 \times 79 \times 111)+3 \times 32 \times 79$ $\times 111=0$.

Explanation as extracted from the printed page; notation may be imperfect.

Open in whiteboard · Browse this chapter in the app