Question

find the product ab(a2+b2) and evaluate it for a=2 and b=1by2

Answer 1

2(1/2)(2 square +1/2 square)
(4+1/4)
16+1/4
17/4

Answer 2

Answer:

Required product is $8 \frac{1}{2}$

Step-by-step explanation:

Given,

$a = 2 \\ b = \frac{1}{2}$

Now,

$a + b = 2 + \frac{1}{2} \\ = \frac{5}{2}$

and

$ab = 2 \times \frac{1}{2} \\ = 1$

We know,

${a}^{2} + {b}^{2} \\ = {(a + b)}^{2} - 2ab \\ = { (\frac{5}{2}) }^{2} - 2 \times 1 \\ = \frac{25}{4} - 2 \\ = \frac{25 - 8}{2} \\ = \frac{17}{2}$

Now required product,

$ab( {a}^{2} + {b}^{2} ) \\ = 1 \times \frac{17}{2} \\ = \frac{17}{2} \\ = 8 \frac{1}{2}$

This is a problem of Algebra.

Some important Algebra formulas:

${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} \\ {(x + y)}^{3} = {x}^{3} + 3 {x}^{2} y + 3x {y}^{2} + {y}^{3} \\ {(x - y)}^{3} = {x}^{3} - 3 {x}^{2} y + 3x {y}^{2} - {y}^{3} \\ {x}^{3} + {y}^{3} = {(x + y)}^{3} - 3xy(x + y) \\ {x}^{3} - {y}^{3} = {(x - y)}^{3} + 3xy(x - y) \\ {x}^{2} - {y}^{2} = (x + y)(x - y) \\ {x}^{2} + {y}^{2} = {(x - y)}^{2} + 2xy \\ {x}^{2} - {y}^{2} = {(x + y)}^{2} - 2xy \\ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} ) \\ {x}^{3} + {y}^{3} = (x - + y)( {x}^{2} - xy + {y}^{2} )$

Know more about Algebra,

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