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प्रश्न
If a, b, c, d are in proportion, then prove that
`[a^2 + ab + b^2]/[a^2 - ab + b^2] = [c^2 + cd + d^2 ]/[ c^2 - cd + d^2 ]`
उत्तर
It is given that a, b, c, d are in proportion.
`therefore a/b = c/d = k`
⇒ a = bk , c = dk
LHS = `[a^2 + ab + b^2]/[a^2 - ab + b^2]`
= `[(bk)^2 + bk xx b + b^2]/[(bk)^2 - bk xx b + b^2]`
= `[b^2(k^2 + k + 1)]/[b^2(k^2 + k + 1 )]`
= `[ k^2+k+1 ]/[ k^2 - k +1 ]` ....(1)
RHS = `[c^2 + cd + d^2 ]/[ c^2 - cd + d^2 ]`
= `[(dk)^2 + dk xx d + d^2]/[(dk)^2 - dk xx d + d^2 ]`
= `[d^2(k^2 + k + 1)]/[d^2(k^2 - k + 1 )]`
= `[k^2+k+1]/[k^2 - k +1]` ...(2)
LHS = RHS
From (1) and (2), we get
`[a^2 + ab + b^2]/[a^2 - ab + b^2] = [c^2 + cd + d^2 ]/[ c^2 - cd + d^2 ]`
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