find the cube of 5 is to 2 by 7
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Since, other solution have described the procedure nicely,I will just briefly go through that and then add something extra.
1)call each term as a or b
2) Find a3a3,b3b3 and abab
3)Use (a+b)3(a+b)3 =a3+b3+3ab(a+b)a3+b3+3ab(a+b) identity.
You know everything in the last equation,except a+b which you can solve and find.
EXTRA:
One notices by observation that (7+52–√)1/3=1+2–√(7+52)1/3=1+2
(7−52–√)1/3=1−2–√(7−52)1/3=1−2
Hence the sum =(1+2–√)+(1−2–√)(1+2)+(1−2)
=22
Hence solved.
1)call each term as a or b
2) Find a3a3,b3b3 and abab
3)Use (a+b)3(a+b)3 =a3+b3+3ab(a+b)a3+b3+3ab(a+b) identity.
You know everything in the last equation,except a+b which you can solve and find.
EXTRA:
One notices by observation that (7+52–√)1/3=1+2–√(7+52)1/3=1+2
(7−52–√)1/3=1−2–√(7−52)1/3=1−2
Hence the sum =(1+2–√)+(1−2–√)(1+2)+(1−2)
=22
Hence solved.
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