Find (a + b)4 -(a - b)4. Hence, or otherwise evaluate ( 3+ 2 )- ( 3- 2 )

(a+b)^4-(a-b)^4 = [ ^4C _0a^4b^0+ ^4C _1a^3b^1+ ^4C _2a^2b^2 C _3a^1b^3+ ^4C _4a^0b^4 ] - [ ^4C _0a^4b^0+ ^4C _1a^3b^1+ ^4C _2a^2b^2- ^4C _3a^1b^3+ ^4C _4a^0b^4 ] = [ ^4C _0a^4(-b)^0+ ^4C _1a^3(-b)^1+ ^4C _2a^2(-b)+ ^4C _3a^1(-b)^3+ ^4C _4a^0(-b)^4 ] - [ ^4C _0a^4(-b)^0+ ^4C _1a^3(-b)^1+ ^4C _2a^2(-b)^2+ ^4C _3a^1(-b)^3+ ^4C _4a^0(-b)^4 ] = [ ^4C _0a^4+ ^4C _1a^3b+ ^4C _2a^2b^2+ ^4C _3ab^3+ ^4C _4ab^4 ] - [ ^4C _0a^4- ^4C _1a^3b+ ^4C _2a^2b^2- ^4C _3ab^3+ ^4C _4b^4 = ^4C _0a^4+ ^4C _1a^3b+ ^4C _2a^2b^2+ ^4C _3ab^3+ ^4C _4ab^4 - ^4C _0a^4+ ^4C _1a^3b- ^4C _2a^2b^2+ ^4C _3ab^3+ ^4C _4b^4 ] =2 [ ^4C _1a^3b+ ^4C _3ab^3 ] =2 [4a^3b+4ab^3 ] =8 [a^3b+ab^3 ] (a+b)^4-(a-b)^4=8(a^3b+ab^3) Putting a= 3 and b= 2 in equation (i) we get ( 3+ 2)^4-( 3- 2)^4=8 [( 3)^3 2+( 3) ( 2)^2 ] =8 [3 6+2 6 ] =8 5 6 =40 6 (3+ 2)^4-(3- 2)^440 6.

Find (a + b)4 -(a - b)4. Hence, or otherwise evaluate ( 3+ 2 )- (…

Download App

Question

Find (a + b)⁴ -(a - b)⁴. Hence, or otherwise evaluate $\Big(\sqrt3+\sqrt2\Big)-\Big(\sqrt3-\sqrt2\Big)$

✓

Answer

$(\text{a}+\text{b})^4-(\text{a}-\text{b})^4$

$=\big[{^4\text{C}}_0\text{a}^4\text{b}^0+{^4\text{C}}_1\text{a}^3\text{b}^1+{^4\text{C}}_2\text{a}^2\text{b}^2{\text{C}}_3\text{a}^1\text{b}^3+{^4\text{C}}_4\text{a}^0\text{b}^4\big]\\-\big[{^4\text{C}}_0\text{a}^4\text{b}^0+{^4\text{C}}_1\text{a}^3\text{b}^1+{^4\text{C}}_2\text{a}^2\text{b}^2-{^4\text{C}}_3\text{a}^1\text{b}^3+{^4\text{C}}_4\text{a}^0\text{b}^4\big]$

$=\big[{^4\text{C}}_0\text{a}^4(-\text{b})^0+{^4\text{C}}_1\text{a}^3(-\text{b})^1+{^4\text{C}}_2\text{a}^2(-\text{b})+{^4\text{C}}_3\text{a}^1(-\text{b})^3+{^4\text{C}}_4\text{a}^0(-\text{b})^4\big]\\-\big[{^4\text{C}}_0\text{a}^4(-\text{b})^0+{^4\text{C}}_1\text{a}^3(-\text{b})^1+{^4\text{C}}_2\text{a}^2(-\text{b})^2+{^4\text{C}}_3\text{a}^1(-\text{b})^3+{^4\text{C}}_4\text{a}^0(-\text{b})^4\big]$

$=\big[{^4\text{C}}_0\text{a}^4+{^4\text{C}}_1\text{a}^3\text{b}+{^4\text{C}}_2\text{a}^2\text{b}^2+{^4\text{C}}_3\text{ab}^3+{^4\text{C}}_4\text{ab}^4\big]\\-\big[{^4\text{C}}_0\text{a}^4-{^4\text{C}}_1\text{a}^3\text{b}+{^4\text{C}}_2\text{a}^2\text{b}^2-{^4\text{C}}_3\text{ab}^3+{^4\text{C}}_4\text{b}^4$

$={^4\text{C}}_0\text{a}^4+{^4\text{C}}_1\text{a}^3\text{b}+{^4\text{C}}_2\text{a}^2\text{b}^2+{^4\text{C}}_3\text{ab}^3+{^4\text{C}}_4\text{ab}^4\\-{^4\text{C}}_0\text{a}^4+{^4\text{C}}_1\text{a}^3\text{b}-{^4\text{C}}_2\text{a}^2\text{b}^2+{^4\text{C}}_3\text{ab}^3+{^4\text{C}}_4\text{b}^4\big]$

$=2\big[{^4\text{C}}_1\text{a}^3\text{b}+{^4\text{C}}_3\text{ab}^3\big]$

$=2\big[4\text{a}^3\text{b}+4\text{ab}^3\big]$

$=8\big[\text{a}^3\text{b}+\text{ab}^3\big]$

$\therefore(\text{a+b})^4-(\text{a-b})^4=8(\text{a}^3\text{b}+\text{ab}^3)$

Putting $\text{a}=\sqrt3$ and $\text{b}=\sqrt2$ in equation (i) we get

$(\sqrt3+\sqrt2)^4-(\sqrt3-\sqrt2)^4=8\Big[(\sqrt3)^3\times\sqrt2+(\sqrt3)\times(\sqrt2)^2\Big]$

$=8\Big[3\sqrt6+2\sqrt6\Big]$

$=8\times5\sqrt6$

$=40\sqrt6$

$\therefore(\sqrt{3}+\sqrt2)^4-(\sqrt{3}-\sqrt2)^440\sqrt6.$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Binomial Theorem→Binomial Theorem — all question types→MATHS STD 11 Science question papers→Rajasthan Board STD 11 Science question papers→Rajasthan Board question paper generator→

Binomial Theorem — MATHS STD 11 Science — Question

Answer

Need a full question paper?

Explore more

Similar questions