问题1.找到以下产品?
一世。 (3x + 2y)(9x 2 – 6xy + 4y 2 )
解决方案:
We know that a3 + b3 = (a + b)(a2 – ab + b2)
we can write the given equation as,
=> (3x + 2y)[(3x)2 – 6xy + (2y)2]
=> (3x)3 + (2y)3
=> 27x3 + 8y3
ii。 (4x – 5y)(16x 2 + 20xy + 25y 2 )
解决方案:
We know that a3 – b3 = (a – b)(a2 + ab + b2)
we can write the given equation as,
=> (4x – 5y)[(4x)2 + 20xy + (5y)2]
=> (4x)3 – (5y)3
=> 64x3 – 125y3
iii。 (7p 4 + q)(49p 8 – 7p 4 q + q 2 )
解决方案:
We can write the given equation as,
=> (7p4 + q)[(7p4)2 – 7p4q + q2]
We know that a3 + b3= (a + b)(a2 – ab + b2)
=> (7p4)3 + q3
=> 343p12 + q3
iv。 [(X / 2)+ 2Y] [(×2/4) -的xy + 4Y 2]
解决方案:
We can write the given equation as,
[(x / 2) + 2y] [(x / 2)2 – (x / 2) * 2y + (2y)2] —— eq(i)
By writing the given equation as eq(i) we can easily make the equation as (a + b)[a2 – ab + b2] = a3 + b3
So, the above e quation can be solved as,
=> (x / 2)3 + (2y)3
=> (x3 / 8) + 8y3
v。[(3 / x)–(5 / y)] [(9 / x 2 )+(25 / y 2 )+(15 / xy)]
解决方案:
We can write given equation as,
[(3 / x) – (5 / y)] {(3 / x)2 + (3 / x)(5 / y) + (5 / y)2]
So, above equation makes the identity of a3 – b3
Now,
=> (3 / x)3 – (5 / y)3
=> (27 / x3) – (125 / y3)
vi。 [3 +(5 / x)] [9 –(15 / x)+(25 / x 2 )]
解决方案:
We can write the given equation as,
=> [3 + (5 / x)] [(3)2 – 3 * (5 / x) + (5 / x)2]
So, above equation makes the identity of a3 + b3
=> (3)3 + (5 / x)3
=> 27 + (125 / x2)
七。 [(2 / x)+ 3x] [(4 / x 2 )+ 9x 2 – 6)]
解决方案:
We can write the given equation as,
=> [(2 / x) + 3x] [(2 / x)2 – (2 / x)(3x) + (3x)2]
So, above equation makes the identity of a3 + b3
=> (2 / x)3 + (3x)3
=> (8 / x3) + 27x3
八。 [(3/2)– 2x 2 ] [(9 / x 2 )+ 4x 4 – 6x]
解决方案:
We can write the given equation as,
=> [(3 / x) – 2x2] [(3 / x)2 – (3 / x)(2x2) + (2x2)2]
So, above equation makes the identity of a3 – b3
=> (3 / x)3 – (2x2)3
=> (27 / x3) – 8x6
ix。 (1 – x)(1 + x + x 2 )
解决方案:
This equation is clearly making the identity of a3 – b3
=> 13 – x3
=> 1 – x3
X。 (1 + x)(1 – x + x 2 )
解决方案:
This equation is clearly making the identity of a3 + b3
=> 13 + x3
=> 1 + x3
xi (x 2 – 1)(x 4 + x 2 +1)
解决方案:
We can write the given equation as,
=> (x2 – 1 ) [(x2)2 + x2 + 1)]
This equation is clearly making the identity of a3 – b3
=> (x2)3 – 1
=> x6 – 1
十二。 (x 3 +1)(x 6 – x 3 +1)
解决方案:
We can write the given equation as,
=> (x3 + 1) [(x3)2 – x3 + 1]
This equation is clearly making the identity of a3 + b3
=> (x3)3 + 1
=> x9 + 1
问题2.如果x = 3且y = -1,请使用indentity查找以下各项的值?
一世。 (9y 2 – 4x 2 )(81y 4 + 36x 2 y 2 + 16x 4 )
解决方案:
We can write the given equation as,
=> (9y2 – 4x2) [(9y2)2 + 9y2 * 4x2 + (4x2)2]
This is now clearly making the identity of a3 – b3
=> (9y2)3 – (4x2)3
=> 729y6 – 64x6 —–eq(i)
Putting the given values in eq(i)
=> 729 * 1 – 64 * 729
=> 729 – 46656
=> -45927
ii。 [(3 / X) – (X / 3)] [(×2/9)+(9 /×2)+ 1]
解决方案:
We can write the given equation as,
=> [(3 / x) – (x / 3)] [(x / 3)2 + (x / 3)(3 / x) + (3 / x)2]
This is making the identity of a3 – b3
=> (3 / x)3 – (x / 3)3 —-eq(i)
Putting the given values in eq(i)
=> 1 – 1
=> 0
iii。 [(X / 7)+(Y / 3)] [(×2/49)+(Y 2/9) – (XY / 21)]
解决方案:
We can write the given equation as,
=> [(x / 7) + (y / 3)] [(x / 7)2 – (x / 7)(y / 3) – (y / 3 )2]
This is making the identity of a3 + b3
=> (x / 7)3 + (y / 3)3 —eq(i)
Putting the values in eq(i)
=> 27 / 343 – 1 / 27
=> (729 – 343) / 9261
=> 386 / 9261
iv。 [(X / 4) – (Y / 3)] [(×2/16)+(XY / 12)+(Y 2/9)]
解决方案:
We can write this equation as,
=> [(x / 4) – (y / 3)] [(x / 4)2 + (x / 4)(y / 3) + (y / 3)2]
This is clearly making the identity of a3 – b3
=> (x / 4)3 – (y / 3)3
=> (x3 / 64) – (y3 / 27) —eq(i)
Putting the values in eq(i)
=> (27 / 64) + (1 / 27)
=> (729 + 64) / 1728
=> 793 / 1728
问题3.如果a + b = 10且ab = 16,求a 2 -ab + b 2和a 2 + ab + b 2的值?
解决方案:
Taking a + b = 10
On squaring both sides,
=> (a + b)2 = (10)2
We get, a2 + b2 + 2ab = 100 —eq(i)
Putting the value of ab = 16 in eq(i)
=> a2 + b2 + 2 * 16 = 100
=> a2 + b2 + 32 =100
=> a2 + b2 = 100 – 32 = 68
So, a2 – ab + b2 = a2 + b2 – ab = 68 – 16 = 52
and a2 + ab + b2 = a2 + b2 + ab = 68 + 16 = 84
问题4.如果a + b = 8且ab = 6,求a 3 + b 3的值?
解决方案:
Taking a + b = 8
On cubing both sides,
(a + b)3 = (8)3
=> a3 + b3 + 3ab(a + b) = 512 —-eq(i)
Putting the given values in eq(i)
=> a3 + b3 + 3 * 6 * 8 = 512
=> a3 + b3 + 144 = 512
=> a3 + b3 = 512 – 144 = 368
=> a3 + b3 = 368
问题5.如果a – b = 6且ab = 20,求a 3 – b 3的值?
解决方案:
Taking a – b=6
On cubing both sides,
(a – b)3 = (6)3
=> a3 – b3 – 3ab(a – b) = 216 —eq(i)
Putting the given values in eq(i)
=> a3 – b3 – 3 * 20 * 6 = 216
=> a3 – b3 – 360 = 216
=> a3 – b3 = 216 + 360 = 576
=> a3 – b3 = 576
问题6.如果x = -2且y = 1,则使用一个标识找到以下值:
一世。 (4y 2 – 9x 2 )(16y 4 + 36x 2 y 2 + 81x 4 )
解决方案:
Given equation can be written as,
=> (4y2 – 9x2)[(4y2)2 + 4y2 * 9x2 + (9x2)2]
This equation now making the identity of a3 – b3
=> (4y2)3 – (9x2)3
=> 64y6 – 729x6 —–eq(i)
Putting the given values in eq(i)
=> 64 * 16 – 729 * (-2)6
=> 64 – 729 * 64
=> 64 – 46656
=> -46592
ii。 [(2 / X) – (X / 2)] [(4 /×2)+(X 2/4)+ 1]
解决方案:
We can write this equation as,
=> [(2 / x) – (x / 2)] [(2 / x)2 + 2(2 / x)(x / 2) + (x / 2)2]
This equation is clearly making the identity of a3 – b3
=> (2 / x)3 – (x / 2)3
=> (8 / x3) – (x3 / 8) —eq(i)
Putting the given values in eq(i)
=> [8 / (-2)3] – [(-2)3 / 8]
=> -1 + 1
=> 0