GIVEN: [42+27+44+213+48+2x] FORMULA USED: (x+y+z)2=x2+y2+z2+2xy+2yz+2zx CALCULATION: [42+27+44+213+48+2x] Changing all the terms into the power of ‘2’: ⇒[24+27+28+213+216+2x] ⇒[24+28+216+27+213+2x] ⇒[24+28+216+2×26+2×212+2×2(x−1)] Here, the expression inside the bracket will be perfect square of [22+24+28] And [22+24+28]2=[(22)2+(24)2+(28)2+2×22×24+ 2×24×28+2×22×28]=[24+28+216+2×26+ 2×212+2×210] Comparing this with the given expression: (x – 1) = 10 ⇒ x = 11