Question

plz prove it.plz plz plz.​

Answer 1

Step-by-step explanation:

${x}^{ {a}^{2} {b}^{ - 1} {c}^{ - 1} } .{x}^{ {b}^{2} {a}^{ - 1} {c}^{ - 1} }. {x}^{ {c}^{2} {a}^{ - 1} {b}^{ - 1} } = {x}^{2} \\ LHS = {x}^{ {a}^{2} {b}^{ - 1} {c}^{ - 1} } .{x}^{ {b}^{2} {a}^{ - 1} {c}^{ - 1} }. {x}^{ {c}^{2} {a}^{ - 1} {b}^{ - 1} } \\ = {x}^{ \frac{ {a}^{2} }{bc} } .{x}^{ \frac{ {b}^{2} }{ac} } .{x}^{ \frac{ {c}^{2} }{ab} } \\ = {x}^{ \frac{ {a \times a}^{2} }{a \times bc} } .{x}^{ \frac{ {b \times b}^{2} }{b \times ac} } .{x}^{ \frac{ {c \times c}^{2} }{c \times ab} } \\ = {x}^{ \frac{ {a}^{3} }{abc} } .{x}^{ \frac{ {b}^{3} }{abc} } .{x}^{ \frac{ {c}^{3} }{abc} } \\ = {x}^{\frac{ {a}^{3} }{abc} +\frac{{b}^{3} }{abc} + \frac{{c}^{3} }{abc} } \\ = {x}^{ \frac{{a}^{3} + {b}^{3} + {c}^{3} }{abc} } \\ = {x}^{ \frac{2({a}^{3} + {b}^{3} + {c}^{3}) }{2abc} } \\ ( \because \: given\: {a}^{3} + {b}^{3} + {c}^{3} = 1 \: \\ and \: 2abc = 1)\\ = {x}^{ \frac{2(1) }{1} } \\ = {x}^{ \frac{2}{1} } \\ = {x}^{2}$

= RHS

HENCE PROVED