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Find (a + b)4 – (a – b)4. Hence, evaluate (3+2)4-(3-2)4 - Mathematics

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प्रश्न

Find (a + b)4 – (a – b)4. Hence, evaluate `(sqrt3 + sqrt2)^4 - (sqrt3 - sqrt2)^4`

योग

उत्तर

Using Binomial Theorem, the expressions, (a + b)4 and (a – b)4, can be expanded as

`(a + b)^4  =  ^4C_0  a^4  +  ^4C_1  a^3  b  +  ^4C_2   a^2b^2  +  ^4C_3  ab^3  + ^4C_4  b^4`

(a - b)4 = 4C0 a4 - 4C1 a3b + 4C2 a2b2 - 4C3 ab3 + 4C4b

∴ `(a + b)^4 - (a - b)^4 =  ^4C_0  a^4  +  ^4C_1  a^3  b  +  ^4C_2   a^2b^2  +  ^4C_3  ab^3  +  ^4C_4  b^4`

[4C0 a4 - 4C1 a3b + 4C2 a2b2 - 4C3 ab3 + 4C4 b4]

2 (4C1a3b + 4C3ab3) = 2(4a3b + 4ab3)

= 8ab (a2 + b2)

In this, by substituting `a = sqrt 3 , b = sqrt 2`

`(sqrt3  +  sqrt2)^4  - (sqrt3  -  sqrt2)^4`

= `8sqrt3. sqrt2 [(sqrt3)^2  + (sqrt2)^2]`

= `8sqrt6 (3 + 2)  = 40sqrt6`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Binomial Theorem - Exercise 8.1 [पृष्ठ १६७]

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एनसीईआरटी Mathematics [English] Class 11
अध्याय 8 Binomial Theorem
Exercise 8.1 | Q 11 | पृष्ठ १६७
एनसीईआरटी Mathematics [English] Class 11
अध्याय 8 Binomial Theorem
Exercise 8.1 | Q 11 | पृष्ठ १६७

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