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Evaluate (3+2)6-(3-2)6

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प्रश्न

Evaluate `(sqrt3 + sqrt2)^6 - (sqrt3 - sqrt2)^6`

योग
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उत्तर

`(a + b)^6  =  ^6C_0 a^6  +  ^6C_1  a^5 b  +  ^6C_2  a^4  b^2  +  ^6C_3  a^3  b^3  +  ^6C_4  a^2  b^4  +  ^6C_5a^1b^5 +  ^6C_6  b^6`

= `a^6  +  6a^5b  +  15a^4  b^2  + 20a^3  b^3  +  15a^2  b^4  + 6ab^5  +  b^6`

`(a - b)^6 =  ^6C_0 a^6  -  ^6C_1  a^5 b  +  ^6C_2  a^4  b^2  -  ^6C_3  a^3  b^3  +  ^6C_4  a^2  b^4  -  ^6C_5a^1b^5 +  ^6C_6  b^6`

= `a^6  -  6a^5b  +  15a^4  b^2  -  20a^3  b^3  +  15a^2  b^4  - 6ab^5  +  b^6`

∴ `(a + b)^6  - (a -b)^6  =  2(6a^5b  + 20a^3  b^3  + 6ab^5)`

Putting a = `sqrt3` and b = `sqrt2`, we obtain

`(sqrt3 + sqrt2)^6  -  (sqrt3 + sqrt2)^6`  =  `2[6(sqrt3)^5 (sqrt2) + 20 (sqrt3)^3 (sqrt2)^3 + 6 (sqrt3)(sqrt2)^5]`

= `2[54sqrt6 + 120 sqrt6 + 24 sqrt6]`

= `2 xx 198 sqrt6`

= `396 sqrt6`

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अध्याय 7: Binomial Theorem - Miscellaneous Exercise [पृष्ठ १३३]

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एनसीईआरटी Mathematics [English] Class 11
अध्याय 7 Binomial Theorem
Miscellaneous Exercise | Q 2. | पृष्ठ १३३

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