Question

solve the problem for class 9

Answer 1

Question :-

$(x {}^{a - b} ) {}^{a + b} \times (x {}^{b - c} ) {}^{b + c} \times (x {}^{c - a} ) {}^{c + a} = 1$

Solution :-

Take LHS.

$(x {}^{a - b} ) {}^{a + b} \times (x {}^{b - c} ) {}^{b + c} \times (x {}^{c - a} ) {}^{c + a} \\ \\ apply \: \: (a {}^{m} ) {}^{ {}^{n} } = a {}^{mn} \: \: formula \\ \\ hence , \\ \: \: \\ (x) {}^{a {}^{2} - b {}^{2} } \times (x) {}^{b {}^{2} - c {}^{2} } \times (x) {}^{c {}^{2} - a {}^{2} } \\ \\ now \: \: apply \: \: a {}^{m} \times a {}^{n} = a {}^{m + n} \: \: formula \\ \\ (x) {}^{a {}^{2} - b {}^{2} + b {}^{2} - c {}^{2} + c {}^{2} - a {}^{2} } = x {}^{0} \\ \\ \\ and \: we \: know \: that \: if \: a \: \\ number \: has \: \: \: power \: 0 \: \: \\ then \: its \: value \: will \: be = 1 \\ \\ so , \: \: \: x {}^{0} = 1 \: \: \: rhs \\ \\ \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: <strong><u>Proved</u></strong><strong><u> </u></strong>$

Answer 2

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Question}}}}}}}}$

show that:-

$\sf(x^{a - b} )^{a + b} \times (x ^{b - c} )^{b + c} \times (x ^{c - a} )^{c + a} = 1$

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Solution}}}}}}}}$

$\sf(x^{a - b} )^{a + b} \times (x ^{b - c} )^{b + c} \times (x ^{c - a} )^{c + a} = 1$

${\bold{\blue{\boxed{\bf{(a+b)(a-b)=a^2 - b^2}}}}}$

$\sf(x^{a^2 - b^2} ) \times (x ^{b^2 - c^2} ) \times (x ^{c^2 - a^2} ) = 1$

${\bold{\blue{\boxed{\bf{ x^a \times x^b = x^{a+b}}}}}}$

$\sf x^{a^2 - b^2 + b^2 - c^2 + c^2 - a^2}= 1$

$\sf x^{\cancel a^2 - \cancel b^2 + \cancel b^2 - \cancel c^2 + \cancel c^2 - \cancel a^2}= 1$

$\sf x^0 = 1$

${\bold{\blue{\boxed{\bf{x ^0 = 1}}}}}$

$\sf 1 = 1$

............Hence Proved

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