Aptitude - Numbers - Discussion
Discussion Forum : Numbers - General Questions (Q.No. 22)
22.
753 x 753 + 247 x 247 - 753 x 247 | = ? |
753 x 753 x 753 + 247 x 247 x 247 |
Answer: Option
Explanation:
Given Exp. = | (a2 + b2 - ab) | = | 1 | = | 1 | = | 1 |
(a3 + b3) | (a + b) | (753 + 247) | 1000 |
Discussion:
24 comments Page 1 of 3.
Imran khan said:
3 years ago
Let a = 753 and b = 247.
Replace all numbers with "a" and "b".
You will get:
a * a + b * b - a * b/a *a * a + b *b * b,
That is, a^2 + b^2 - ab/a^3 + b^3,
As per formula:
(a^2 + b^2 - ab/a^3 + b^3).
(a^3+b^3)=(a+b)(a^2+b^2-ab),
= (a^2+b^2-ab)/(a^3+b^3),
= (a^2+b^2-ab)/(a+b)(a^2+b^2-ab).
= 1/(a+b).
1/(743+247) => 1/1000.
Replace all numbers with "a" and "b".
You will get:
a * a + b * b - a * b/a *a * a + b *b * b,
That is, a^2 + b^2 - ab/a^3 + b^3,
As per formula:
(a^2 + b^2 - ab/a^3 + b^3).
(a^3+b^3)=(a+b)(a^2+b^2-ab),
= (a^2+b^2-ab)/(a^3+b^3),
= (a^2+b^2-ab)/(a+b)(a^2+b^2-ab).
= 1/(a+b).
1/(743+247) => 1/1000.
(5)
Chaitra said:
4 years ago
Formula:(a^3+b^3)=(a+b)(a^2+b^2-ab),
= (a^2+b^2-ab)/(a^3+b^3),
= (a^2+b^2-ab)/(a+b)(a^2+b^2-ab).
= 1/(a+b).
= (a^2+b^2-ab)/(a^3+b^3),
= (a^2+b^2-ab)/(a+b)(a^2+b^2-ab).
= 1/(a+b).
(4)
Jyoti said:
5 years ago
I didn't understand please solve in a simple way.
(3)
Thula sunya behuria said:
5 years ago
I am not understanding it. Please explain it.
(2)
Shipra said:
6 years ago
(753) ^2+(247) ^2-753*247/(753) ^3+(247) ^3.
= (753) +(247) /+(753) ^3+(247) ^3.
= 1/(753) ^2+(247) ^2.
= 1/(753+247).
= 1/1000.
= (753) +(247) /+(753) ^3+(247) ^3.
= 1/(753) ^2+(247) ^2.
= 1/(753+247).
= 1/1000.
(1)
Kiran said:
6 years ago
Can you explain the formula for the answer?
Zain said:
6 years ago
I am not getting this, Please anyone explain me.
Arjuna said:
6 years ago
Hi, can anyone explain it in the direct method?
(1)
Arjuna said:
6 years ago
Here, we can use.
(a^3+b^3) = (a+b) (a^2-ab+b^2).
1/(a+b) = (a^2-ab+b^2)/(a^3+b^3).
(a^3+b^3) = (a+b) (a^2-ab+b^2).
1/(a+b) = (a^2-ab+b^2)/(a^3+b^3).
(4)
Sriji said:
8 years ago
Explain the formula please.
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