1.यदि `x-\frac{1}{x}, तब x^3-\frac{1}{x^3}` किसके बराबर है ?
solution -
given thant `x-\frac{1}{x}= 6`
दोनों और घन करने पर
`(x-\frac{1}{x})^3 = 6^3`
`x^3-\frac{1}{x^3} -3\times x\times\frac{1}{x} (x-\frac{1}{x})= 216` [ SSC CPO SI 2019]
`x^3-\frac{1}{x^3}-3(x-\frac{1}{x})= 216`
`x^3-\frac{1}{x^3}-3\times6= 216`
`x^3-\frac{1}{x^3}-18= 216`
`x^3-\frac{1}{x^3}= 216+18`
`x^3-\frac{1}{x^3}= 234`
2. यदि `a^3+b^3=5824` और `a+b=28` तो `(a-b)^2+a\times b` = ?
given `a^3+b^3=5824` और `a+b=28`
`a^3+b^3=5824` ----------(1)
`a+b=28` -------------(2)
on taking cubes bothsides we have
`(a+b)^3 = 28^3`
`a^3+b^3+3 \times a\times b \times (a+b) = 21952` from equation 1
`5824+3\times ab \times 28 = 21952` from equation 2
`84\times ab = 21952-5824`
`84\times ab =16128`
`ab = 192`----------------(3)
now squaring equation no 2 we have
`(a+b)^2 = 28^2`
`a^2+b^2+2\times ab = 784`
`a^2+b^2+2\times 192 = 784`
`a^2+b^2+384 = 784`
`a^2+b^2 = 784-384`
`a^2+b^2= 400`--------------(4)
now
`(a-b)^2+ab`
`= a^2+b^2- 2\times ab + ab`
`= a^2+b^2- ab` from equation 3 & 4
`=400-192`
`=208`
Tags
Mathematics