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1.Prove ( a + b ) ( a − b ) ≡ a 2 − b 2 {\displaystyle {\begin{smallmatrix}(a+b)(a-b)\equiv a^{2}-b^{2}\end{smallmatrix}}} L . H . S . = ( a + b ) ( a − b ) = a ( a − b ) + a ( a − b ) = a 2 + a b − b a − b 2 = a 2 − b 2 R . H . S . = a 2 − b 2 L . H . S . ≡ R . H . S . ∴ I t i s a n i d e n t i t y . {\displaystyle {\begin{aligned}L.H.S.&=(a+b)(a-b)\\&=a(a-b)+a(a-b)\\&=a^{2}+ab-ba-b^{2}\\&=a^{2}-b^{2}\\R.H.S.&=a^{2}-b^{2}\\L.H.S.&\equiv R.H.S.\\\therefore It\;is\;an\;identity.\\\end{aligned}}}
2.If A x 2 + B x B x + 6 C x 2 − 9 ≡ ( B + C ) x 2 + ( C x + 2 x ) B x + 6 5 x 2 − 9 {\displaystyle {\begin{smallmatrix}Ax^{2}+Bx{\frac {Bx+6}{Cx^{2}-9}}\equiv (B+C)x^{2}+(Cx+2x){\frac {Bx+6}{5x^{2}-9}}\end{smallmatrix}}} , find the Constants A,B and C L . H . S . = A x 2 + B x B x + 6 C x 2 − 9 R . H . S . = ( B + C ) x 2 + ( C x + 2 x ) B x + 6 5 x 2 − 9 = ( B + C ) x 2 + ( C + 2 ) x B x + 6 5 x 2 − 9 {\displaystyle {\begin{aligned}L.H.S.&=Ax^{2}+Bx{\frac {Bx+6}{Cx^{2}-9}}\\R.H.S.&=(B+C)x^{2}+(Cx+2x){\frac {Bx+6}{5x^{2}-9}}\\&=(B+C)x^{2}+(C+2)x{\frac {Bx+6}{5x^{2}-9}}\\\end{aligned}}} ∴ A x 2 + B x B x + 6 C x 2 − 9 ≡ ( B + C ) x 2 + ( C + 2 ) x B x + 6 5 x 2 − 9 {\displaystyle \therefore Ax^{2}+Bx{\frac {Bx+6}{Cx^{2}-9}}\equiv (B+C)x^{2}+(C+2)x{\frac {Bx+6}{5x^{2}-9}}} C = 5 B = 5 + 2 = 8 A = 8 + 5 = 13 {\displaystyle {\begin{aligned}C&=5\\B&=5+2\\&=8\\A&=8+5\\&=13\\\end{aligned}}}
