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# 1. Utilizing (a-b)3 = a3-b3-3ab(a-b), tick the right expression: A. X3+3x=p+(1/p) B.x3+3x=p-(1/p) C.x3+3x=p+1 2. On the off chance that a = 31/4 + 3-1/4 and b=31/4 - 3-1/4 , then the estimation of (a2 + b2)2 is … 3. On the off chance that a = ( 21/2 + )1/3 - (21/2 – 1)1/3 , then the estimation of a3 + 3a – 2 is … . 4. On improvement ( xa/(a-b)/xa/(a+b) ) ÷ (xb/(b-a)/xb/(b+a)) ) a+b decreases to …

1. Utilizing (a-b)3 = a3-b3-3ab(a-b), tick the right expression: A. X3+3x=p+(1/p) B.x3+3x=p-(1/p) C.x3+3x=p+1 2. On the off chance that a = 31/4 + 3-1/4 and b=31/4 - 3-1/4 , then the estimation of (a2 + b2)2 is … 3. On the off chance that a = ( 21/2 + )1/3 - (21/2 – 1)1/3 , then the estimation of a3 + 3a – 2 is … . 4. On improvement ( xa/(a-b)/xa/(a+b) ) ÷ (xb/(b-a)/xb/(b+a)) ) a+b decreases to …

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1. b 2.20/3

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3.a = 1 therefore,a^3+3a+-2 = 1+3-2 = 2 4. is a recurrent improvement so it diminishes 1.

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3. a + (2^0.5 - 1 )^(1/3 ) = (2^0.5 +1)^(1/3) in the event that we block on both sides we get a^3 + 2^0.5 - 1 +3*a*(2^0.5 - 1 )^(1/3 )*(a + (2^0.5 - 1 )^(1/3 )) = 2^0.5 + 1 suggests a^3 + 2^0.5 - 1 +3*a*(2^0.5 - 1 )^(1/3 )*((2^0.5 +1)^(1/3)) = 2^0.5 + 1 infers a^3 +3a - 2 = 0; 4. numerator:x^(a/(a-b) - an/(a+b)) = x^(2ab/(a^2-b^2)) likewise denominator is x^(2ab/(b^2 - a^2)) so add up to division is x^(4ab/(a^2-b^2)) so part control a+b is x^(4ab/(a-b)) 2. a^2 = 3^0.5 +3^-0.5 +2 b^2 = 3^0.5 + 3^-0.5 - 2 a^2 + b^2 = 2(3^0.5 + 3^-0.5) (a^2 + b^2)^2 = 4(3 + 1/3 + 2) = 64/3

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