Three numbers are to one another 2:3:4. The sum of their cubes is 33957. Find the largest number.
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Answered by
6
Let the numbers be 2x,3x,4x.............1
according to problem:
[2x]³+[3x]³+[4x]³=33957
⇒8x³+27x³+64x³=33957
⇒99x³=33957
⇒x³=343
⇒x=7
putting x=7 in 1,we get
2×7=14
3×7=21
4×7=28
according to problem:
[2x]³+[3x]³+[4x]³=33957
⇒8x³+27x³+64x³=33957
⇒99x³=33957
⇒x³=343
⇒x=7
putting x=7 in 1,we get
2×7=14
3×7=21
4×7=28
Lokeshk:
That's very sweet of you. thanks
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Answered by
2
Let the 3 nos. be 2x, 3x, and 4x
so, (2x)^3 + (3x)^3 + (4x)^3 = 33957
=> 8x^3 + 27x^3 + 64x^3 = 33957
=> 99x^3 = 33957
=> x^3 = 33957/99 = 343
=> x = cube root of 343 = 7
Now, we have found that value of x = 7 so the value of the largest number is 4x = 4*7 = 28
so, (2x)^3 + (3x)^3 + (4x)^3 = 33957
=> 8x^3 + 27x^3 + 64x^3 = 33957
=> 99x^3 = 33957
=> x^3 = 33957/99 = 343
=> x = cube root of 343 = 7
Now, we have found that value of x = 7 so the value of the largest number is 4x = 4*7 = 28
thanks
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