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:
4 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.
(10)
Nirmal Saxena said:
9 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) = 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) = 1/(a + b),
1/(743+247) => 1/1000.
(1)
P.s.Deepa said:
1 decade ago
In the denominator
=(753*753*753)+(247*247*247)
=(a*a*a)+(b*b*b)
=a3+b3
Here we assume 753 as "a" and 247 as "b". Since in the denominator 753 is three times so we are writing a^3 and 247 is three times so b^3.
=(753*753*753)+(247*247*247)
=(a*a*a)+(b*b*b)
=a3+b3
Here we assume 753 as "a" and 247 as "b". Since in the denominator 753 is three times so we are writing a^3 and 247 is three times so b^3.
Sangeetha said:
10 years ago
(753)^2+(247)^2-(753)(247)
------------------------------------
(753) ^3 + (247) ^3
a^2+b^2-ab.
=------------------.
a^3 + b^3.
Since, a^3+b^3 = (a+b) (a^2 + b^2 - ab).
Therefore, 1/(a+b) is the answer (1/1000).
------------------------------------
(753) ^3 + (247) ^3
a^2+b^2-ab.
=------------------.
a^3 + b^3.
Since, a^3+b^3 = (a+b) (a^2 + b^2 - ab).
Therefore, 1/(a+b) is the answer (1/1000).
Shweta said:
9 years ago
@Divi.
(a^2 + b^2 - ab)/(a^3 + b^3).
Formula: (a^3 + b^3) = (a+b)(a^2 + b^2 - ab).
(a^2 + b^2 - ab)/(a + b)(a^2 + b^2 - ab) = 1/(a+b).
(a^2 + b^2 - ab)/(a^3 + b^3).
Formula: (a^3 + b^3) = (a+b)(a^2 + b^2 - ab).
(a^2 + b^2 - ab)/(a + b)(a^2 + b^2 - ab) = 1/(a+b).
(1)
Abhishek said:
1 decade ago
Pritika, its simple. Look.
a3 + b3= (a + b) (a2 + b2 - ab). It is maths formula.
Therefore put the formula. You get the answer.
a3 + b3= (a + b) (a2 + b2 - ab). It is maths formula.
Therefore put the formula. You get the answer.
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.
(2)
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).
(5)
Muhammad Ahsan said:
9 years ago
Can anybody explain after this step?
a2+b2-ab / a2+b2-ab.
How 1/a+b comes from that above equation?
a2+b2-ab / a2+b2-ab.
How 1/a+b comes from that above equation?
Arjuna said:
7 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)
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